Confidence Intervals for Population Mean: Videos & Practice Problems

To construct a confidence interval for a population mean, use the sample mean as a point estimate and calculate the margin of error with the equation $e_m=z_{\mathrm{\alpha /2}}\times \frac{\sigma }{\sqrt{n}}.$ If the population standard deviation is unknown, substitute the sample standard deviation and use the t distribution. Ensure the sample is random and meets normality conditions. Interpret the interval to express confidence in the population mean's range.

Population Standard Deviation Known

Throughout this chapter, we've been taking a look at confidence intervals, and we learned how to find their associated critical z scores. But we have yet to fully construct a confidence interval for a population mean, mu. So that's exactly what we're going to look at here, putting everything that we've learned so far together in a step-by-step process to ultimately construct a confidence interval for you. So let's go ahead and get started here by jumping right into our example. Here, we're told that over 36 trips to work, you find a sample mean of one hour.

Now we're asked here to construct a 90% confidence interval for the true population mean travel time, and we're told here to use a population standard deviation of eighteen minutes. Now, since we're explicitly given our population standard deviation sigma in this problem, that then means that we can use x bar, our sample mean, as a point estimate and calculate our margin of error using this equation: \( e = z_{\alpha/2} \times \frac{\sigma}{\sqrt{n}} \). So we can ultimately use this to construct our confidence interval. But before we get there, we have to start from the beginning with step one.

We first need to verify that we can even follow this process. The first thing that we need to check for here is that our sample is random. Now in most of these problems, you can assume that your sample is random unless you're told that it's not explicitly. In this problem, all we're told is that this is over 36 trips to work. We have no reason to think that the sample isn't random.

So we can go ahead and check that box off. Now the second thing that we want to verify is that x is normal or that our sample size n is greater than 30. This is ultimately to check that this particular random variable has a standard normal distribution. Now here we're told that our sample size n is 36. Since this value is greater than 30, we can check that box off and move on to step two, finding our critical value z alpha over two.

In order to find z alpha over two, we first need to figure out what alpha over two is. Remember that alpha is equal to one minus c, our confidence level. So we're just taking that and dividing it by two. Now in this problem, we have a 90% confidence level. So this becomes one minus 0.9 over two, which gives us a value here of 0.05.

Now from here, you can either use a calculator or a z table to look up what our critical z value is with this alpha over two value of 0.05. Whatever way you choose to do this, you should end up finding that z alpha over two is equal to 1.645 for this 90% confidence level. So with step two done, we can move on to step three and actually find our margin of error. This is using the equation that we just learned up there. This is \( z_{\alpha/2} \times \frac{\sigma}{\sqrt{n}} \).

So let's go ahead and plug all of those values in to find our margin of error. First, taking that critical z value we just found, 1.645, then we're multiplying that by that population standard deviation 18 over the square root of our sample size, and that is 36. Now when we multiply this out, we will ultimately get that our margin of error here is equal to 4.935. So that's step three done. Moving on to our final step here, we're actually going to construct that confidence interval.

We're going to take our sample mean here x bar and subtract our margin of error for that lower bound and add our margin of error for that upper bound. So constructing our confidence interval here, abbreviating this CI for confidence intervals, I'm going to take my sample mean of one hour. Remember, this is equivalent to sixty minutes. So taking that sixty minutes and subtracting my margin of error 4.935 and then adding 4.935 to 60 for that upper bound. Subtracting and adding here, this ends up giving me 55.065 for that lower bound and 64.935 for that upper bound.

What does this actually mean? Well, given our confidence level of 90% and this range of values from 55.065 to 64.935, we can then interpret this as saying that we are 90% confident based on that confidence level that the mean travel time to work is between these two values fifty-five point zero six five minutes and sixty-four point nine three five minutes. So that is our fully constructed and interpreted confidence interval. Now we're going to get some more practice coming up next. And if you have any questions, feel free to let us know.

I'll see you in the next video.

Population Standard Deviation Known Example 1

If you haven't already given this problem a try on your own, go ahead and pause here and do that before checking back in with me. In this problem, we're told that we consider how much time it takes to find a parking spot and find a mean of one hour and twenty-two minutes with a population standard deviation of fifteen minutes. We're told to assume here that this information was based on a sample of one hundred different days of circling the parking garage. And we're asked to construct and interpret a 99% confidence interval. Let's get started with our steps here so that we can construct this confidence interval.

The first thing we need to do is verify that our sample is random and that either x is normal or our sample size is greater than 30. Now, as I mentioned before, in most of these problems, you can safely assume that your sample is random unless you're told otherwise. So we can go ahead and check that off. Now for our second thing that we're verifying here, we know that this is over a sample of 100 different days. So this is definitely greater than 30, and we can check that off and move on to step two, finding our critical z value, z alpha over two.

We know that we need to figure out what alpha over two is in order to ultimately find that critical z score. Now here, since we're working with a 99% confidence interval, we need to take one minus that 0.99 and divide that by two. This will ultimately give us an alpha over two value here of 0.005. Now, when we either use a calculator or look this up in a z table, this will result in a critical z value, z alpha over two, of 2.576. So with this critical z value in mind, we can move on to step three, finding our margin of error, e.

This is going to be equal to that critical z value we just found times sigma, our population standard deviation, over the square root of our sample size n. So here, e, our margin of error, is then math xmlns="http://www.w3.org/1998/Math/MathML"> z α / 2 σ 100 * 15. Now typing this into our calculator, this will ultimately give us a margin of error of 3.864. Now that we have that margin of error, we can construct our confidence interval by taking our sample mean, x bar, and subtracting and adding that margin of error. So here, our confidence interval is then equal to that one hour and twenty-two minutes.

That is our sample mean. Now one hour is sixty minutes. If we add that together with twenty-two minutes, that gives us eighty-two minutes. So eighty-two minutes is my sample mean. I am subtracting 3.864 for that lower bound and adding 3.864 for that upper bound. This ultimately results in a confidence interval here of seventy-eight point one three six and eighty-five point eight six four. So this is ultimately my confidence interval. We are asked here to interpret this result. So what actually is going on here? Well, we constructed a 99% confidence interval.

So based on this range of values that we found and that confidence level of 99%, we can say that we are 99% confident that our true population mean time that it takes to find a parking spot is in between the values of seventy-eight point one three six and eighty-five point eight six four. If you have any questions here, feel free to let us know, and I'll see you in the next one.

A retailer wants to estimate the average amount spend by customers on holiday shopping. In a random sample of 50 customers, the average amount spend was $250, and the population standard deviation $\sigma$ is known to be $40. Construct and interpret an 80% confidence interval for the average amount spend by all customers.

• A. (210.00, 290.00)
• B. (248.976, 251.024)
• C. (242.76, 257.24)
• D. (248.72, 251.28)

Answer: (C)

Books get more and more expensive every semester, but the distribution of their prices is always normal. 25 randomly selected students in your school spent, on average $500 with a standard deviation of $50. Construct a 98% confidence interval for the true spending on books.

• A. (476.74, 523.26)
• B. (499.90, 500.10)
• C. (490.20, 509.80)
• D. (488.38, 511.62)

Answer: (A)

You want to purchase one of the new Altima. You randomly select 400 dealerships across the United States and find a mean of $25,000. Assume a population standard deviation of $2500. Construct and interpret a 94% confidence interval for the true mean price for the new Nissan Altima.

• A. (24992.5, 25007.5); We are 94% confident that the true mean price for the new Nissan Altima falls between $24992.5 and $25007.5.
• B. (24882.438, 25117.563); We are 94% confident that the true mean price for the new Nissan Altima falls between $24882.438 and $25117.563.
• C. (24764.875, 25235.15); We are 94% confident that the true mean price for the new Nissan Altima falls between $24764.875 and $25235.15.
• D. (24529.75, 25470.25); We are 94% confident that the true mean price for the new Nissan Altima falls between $24529.75 and $25470.25.

Answer: (C)

Critical Values: t-Distribution

Hey, everyone. We just learned how to create a confidence interval for a population mean, \( \mu \), when we know our population standard deviation, \( \sigma \). But often, we won't actually know what \( \sigma \) is. So how can we create a confidence interval then? Well, whenever \( \sigma \) is unknown, we can actually use our sample standard deviation \( s \) in order to estimate \( \sigma \).

Now in this case, our sampling distribution of \( \overline{x} \), which now relies on \( s \) instead of \( \sigma \), follows what's called a t distribution. Now here, we're going to walk through what a t distribution is and how to find its associated critical values so that we can ultimately construct a confidence interval, whether we know \( \sigma \) or not. So let's go ahead and get started here. Now taking a look at our familiar normal distribution, we know that it has this symmetric bell-shaped curve with a mean of zero and a standard deviation of one. Now taking a look at our t distribution over here, we can see that it has a pretty similar shape to our normal distribution, but it's much shorter and wider.

So it still has a mean of zero, but it has a standard deviation that will always be greater than one because it actually depends on the sample size. Now our t distribution also has an additional parameter to consider called degrees of freedom, which we're ultimately going to use to find our critical t value. Now it's not hugely important that you understand exactly what degrees of freedom are, but it is important that you know that our t distribution has \( n \), our sample size, minus one degrees of freedom. So as our sample size increases, our degrees of freedom increase as well. And here on our curve, we can see that as our degrees of freedom increase from one to 30 to 100, our curve starts to look more and more like the curve of our normal distribution.

So as our degrees of freedom increase, t gets more and more normal. Now the most important thing with our t distribution here is that we're able to find our critical t value, which we can ultimately do by looking it up in a table or by using our calculator the same way we did for our critical z value. But let's get a bit more familiar with this process by working through an example here. Here we're asked to find and compare the critical values for a 95% confidence interval with an alpha over two of 0.025, where \( \sigma \) is known and where \( \sigma \) is unknown, using a sample size of 31. Now starting with part a here, we want to find our critical value to build a confidence interval with a known value of \( \sigma \).

Now whenever we know that population standard deviation, we need to use our critical z value \( z_{\alpha/2} \). Now remember to find \( z_{\alpha/2} \), we just need to know what alpha over two is, which we're given in our problem as 0.025. So coming up here to our z table, remember that we're using that alpha over two, which is in the body of my table here, in order to ultimately find z. So here with this 0.025, I can see that going out to the left and up to the top that this corresponds to a critical z value of negative 1.96. Now remember that this is a negative value because looking at our curve here, we know that this is the area to the left of whatever our boundary is.

So this ultimately gives us negative \( z_{\alpha/2} \). Now that means that our positive \( z_{\alpha/2} \) is then positive 1.96. Now moving on to part b here, we want to find our critical value to build our confidence interval when we don't know what \( \sigma \) is, when \( \sigma \) is unknown. Now whenever we don't know that population standard deviation, we instead need to use our critical t value, \( t_{\alpha/2} \), or you may also see this written as \( t_c \). Now in order to find our critical t value, we still need to know what alpha over two is.

It's the same thing, 0.025, given to us in our problem. But we also need to know one more thing, our degrees of freedom. Remember that for our t distribution, we have \( n - 1 \) degrees of freedom. So here with a sample size of 31, 31 minus one gives me 30 degrees of freedom. So coming up here to my t table, I now need to use alpha over two and my degrees of freedom in order to ultimately find t.

Now because our t table was specifically designed for use with confidence intervals, we can see these alpha over two values across the top here and our degrees of freedom down the side. So we no longer have to worry about this being the area to the left because it's ultimately just the area in either tail, alpha over two, given to us in this table. So here looking across the top for my alpha over two value of that 0.025, I know that my value will then be in this column. And then with 30 degrees of freedom here, we have this row. Now we can see that those intersect right here, giving us a critical t value of \( t = 2.042 \).

Now depending on what t table you're working with, some of them might explicitly give you that confidence level. But remember, it's always a good idea to know what alpha over two is because you might not be given that. That. Now comparing our critical t value to our critical z value, we can see that our t value is greater for the same level of confidence. And this will always be the case because our t distribution is wider and ultimately has a greater margin of error.

Now one final thing that I want to mention here is that we can only use either of these two critical values, \( z_{\alpha/2} \) or \( t_{\alpha/2} \), to construct a confidence interval if our sample is random and either \( x \) is normal, or our sample size \( n \) is greater than 30. So remember to double-check for those things before you proceed in creating a confidence interval using these critical values. Now that we know how to find our critical values, whether we know \( \sigma \) or not, let's continue getting a bit more practice. I'll see you in the next one.

Find the critical value $t_{\frac{\alpha}{2}}$for an 80% confidence interval given a sample size of 51.

• A. 0.10
• B. 1.299
• C. 0.300
• D. 2.598

Answer: (B)

Find the critical value $t_{\frac{\alpha}{2}}$for a 95% confidence interval given a sample size of 6.

• A. 0.025
• B. 5
• C. 1.286
• D. 2.571

Answer: (D)

Critical Values: t-Distribution Example 2

If you haven't already tried this problem on your own, go ahead and do that before checking back in with me. Here we have, for each problem below, determine if you can construct a confidence interval for the population mean, \( \mu \), using the standard normal or the t distribution. Remember that these each correspond to different critical values. We know that we use our critical z value when using the standard normal distribution, and we use our critical t value when using the t distribution. Now, once we figure out if we can construct a confidence interval, we want to know, should we use the standard normal or the t distribution?

And if we cannot, why not? Let's start with scenario eight here. Here, we randomly select 18 test scores from a statistics class. The sample mean of the test scores is 74, and the population standard deviation is six. We're asked to construct a 99% confidence interval for the population mean of the test scores.

Now you can be tempted here to go ahead and jump straight to, okay, I'm given my population standard deviation \( \sigma \). So that means I can use a standard normal distribution, but we can't jump straight there because we need to check for some other things first. One, that our sample is random, which is often something that we can assume unless told anything otherwise. And we also need to check that it is normal or that our sample size is greater than 30. So here I can see that we randomly select.

We're told that explicitly, so I know that my sample is random. And going through this problem, I have this sample size of 18, which is not greater than 30. So that means that this has to be normally distributed in order to actually proceed in creating this confidence interval. And we are not told anything about this being normally distributed. So that means that we cannot use the standard normal or the t distribution because we don't know if our data is normal.

So moving on to part b here or scenario b, we randomly measure the height of 87 men and find an average height of five foot ten with a population standard deviation of five inches. Now assuming that the heights are normally distributed, we're asked to construct a 99% confidence interval for the population mean height. Now again, here, we cannot jump straight to, okay, know the population standard deviation. So I know what distribution to use. We need to double-check for those other things.

Now we're told here that we randomly measure these heights, so our sample is definitely random. And we have a sample size of 87 that is definitely greater than 30. And we don't have to check for anything else here, but we are also told that these heights are normally distributed. So we have a lot to work with here. We know that we can go ahead and proceed in constructing a confidence interval.

This is the point where we wanna look at, okay, do we know \( \sigma \) or not? Because we do know our population standard deviation \( \sigma \), that means that we would then use our normal distribution with a critical value of \( z_{\alpha/2} \) because we know \( \sigma \) and we already verified those other things. Let's move on to our final scenario here, scenario c. Here we have 25 randomly selected students in your school spending an average of $500 on textbooks with a sample standard deviation of $50. Now assuming the spending on books is normally distributed, construct a 98% confidence interval for the population mean textbook spending. Now here, these 25 students are randomly selected, so we know our sample is random.

Because our sample size is 25, that is less than 30. That means that this has to be normally distributed in order to proceed here. And it indeed is normally distributed. So that means that we can look at whether we know \( \sigma \) or not. Now because we know our sample standard deviation, that is \( s \), that means that we cannot use our normal distribution the way that we did in scenario b.

We instead need to use our t distribution and ultimately a critical value of \( t_{\alpha/2} \) because we do not know our population standard deviation. We just know our sample standard deviation. Feel free to let us know if you have any questions here, and let's get a bit more practice with this.

For which of the following scenarios can you NOT create a confidence interval using the standard normal or Student t-distribution?

• A. A real estate agent wants to estimate the average selling price of homes in a neighborhood. She collects random sample data from 8 recent home sales and finds that the sample mean selling price is $350,000 with a sample standard deviation of $25,000. Assuming home prices in the neighborhood are normally distributed, construct a 95% confidence interval for the population mean selling price.
• B. A financial analyst is studying the average return on investment (ROI) for a particular stock. He randomly selects data from 30 investors and finds that the sample mean ROI is 6.2% with a population standard deviation of 1.5%. Construct a 90% confidence interval for the population mean ROI.
• C. An economist is analyzing monthly utility expenses for households. He collects data from 12 households and calculates a sample mean monthly expense of $180 with a sample standard deviation of $15. He also notes that the distribution of utility expenses is heavily skewed to the right. Construct a 90% confidence interval for the population mean monthly utility expense.

Answer: (C)

Population Standard Deviation Unknown

We know how to create a confidence interval for a population mean, mu, when we know our population standard deviation, sigma, by using this equation to calculate our margin of error. But more often than not, we will not know what sigma is. So how do we construct a confidence interval then? Well, it turns out that we're going to do almost an identical calculation, except instead of sigma and z alpha over two, we're just going to use s, our sample standard deviation, and t alpha over two, our critical t value. So with that in mind, let's jump right into our example and follow the exact same steps that we already know.

Here we're told that over 36 trips to work, you find a sample mean of one hour and a sample standard deviation of twenty-one minutes. We're asked here to construct and interpret a 90% confidence interval for the true population mean travel time. So no matter what else is going on in our problem, we always need to start here with step one, verifying that our sample is random and that either the distribution is normal or our sample size, n, is greater than 30. Now our sample being random is often something that we can just assume to be true unless we're told anything otherwise, so we can go ahead and check that off. Now in this problem, we are also told that this is over 36 trips to work.

So, with an n value of 36, we know that we can check this second box as well and move on to step two, finding our critical value. Now if we knew our population standard deviation sigma, we would want to go ahead and use our critical z value here. But because here we're given our sample standard deviation, that's s as opposed to sigma, we cannot use our critical z value and we instead need to use our critical t value, t alpha over two, because this has a t distribution. So, in order to find t alpha over two, we need to know what alpha over two is. Now this is going to be equal to one minus our confidence level of 90%, so that's 0.05.

Now, in order to find this critical t value, we also need to know our degrees of freedom, which remember are equal to n minus one. Now in this case, our n value is 36. So, 36 minus one gives us 35 degrees of freedom. So, with these two values here, we can use them to ultimately find t alpha over two, our critical t value, by either using a calculator or looking this up in a table. Either way, you should find that your critical t value is 1.69.

So, with step two done, we move on to step three, actually finding our margin of error. Now, when we knew our population standard deviation sigma, we found our margin of error by taking our critical z value and multiplying that by sigma over the square root of n. Because now we do not know what sigma is, that means that we need to do this a little bit differently. So our margin of error is instead equal to t alpha over two times s over the square root of n. So all we've done here is replaced z alpha over two with t alpha over two and sigma with s, our sample standard deviation.

Calculating this margin of error, e, is then equal to that t alpha over two, which is 1.69 that we just found in step two, times our sample standard deviation, which is 21, over the square root of our sample size, which here is 36. Now when we multiply this out, this will end up giving us a margin of error of 5.915. Now, from here, we can find our confidence interval by finding our upper and lower bounds, which just means we need to subtract and add our margin of error to our sample mean the same way that we've done before. So, in order to find this confidence interval, I can take my sample mean, which is one hour, which we know is equal to sixty minutes. So, I'm going to take that sixty minutes and subtract 5.915 for that lower bound and also add 5.915 for that upper bound here.

This ultimately results in a lower bound of 54.085 and an upper bound of 65.915. So, this is our confidence interval here. Now, one of the most important parts of constructing confidence intervals is being able to interpret your confidence interval. So what does this actually mean? Well, given this 90% confidence level, we can say that we are 90% confident that the mean time to travel to work is between these two values, fifty-four point zero eight five minutes and sixty-five point nine one five minutes.

So that is our full confidence interval that we both constructed and interpreted. Feel free to let us know if you have any questions. But now that we know how to construct a confidence interval, whether we know sigma or not, let's continue getting a bit more practice. I'll see you in the next video.

You ask 16 people in your Statistics class what their grade is. The data appears to be distributed normally. You find a sample mean and sample standard deviation of 60 and 24, respectively. Construct and interpret a 95% confidence interval for the population mean class grade.

• A. (10.14, 37.86); We are 95% confident that the interval for the population mean class grade falls in between 10.14 and 37.86.
• B. (47.214, 72.786); We are 95% confident that the interval for the population mean class grade falls in between 47.214 and 72.786
• C. (-10.65, 58.65); We are 95% confident that the interval for the population mean class grade falls in between -10.65 and 58.65.
• D. (25.35, 94.65); We are 95% confident that the interval for the population mean class grade falls in between 25.35 and 94.65.

Answer: (B)

Population Standard Deviation Unknown Example 3

Welcome back, everyone. Let's take a look at this example problem here. So we have an HR manager who wants to estimate the average number of sick days taken by employees each year. Now, presumably, this is a big company with lots of employees, so they are going to take a random sample of employees and see if they can estimate the total number of sick days that all of the employees will take. Right.

They take this random sample of employees and they find a couple of key pieces of information here. We find the sample mean is six point eight days. That's the sample mean. In other words, that's your X̄, which is 6.8.

We're going to use this to estimate what is the parameter μ (Mu). What's the population mean? And we have the sample standard deviation of 2.4 days. Remember, that is s, not σ (Sigma). So this s here is 2.4.

All right. So clearly, we can see here that we are trying to find a confidence interval for a population mean, which is μ. But unfortunately, we don't know what σ is. So therefore, we are going to have to follow the steps to find a confidence interval using a t-distribution. So let's go ahead and walk through the steps here.

The first thing we have to do is figure out whether the sample is random or not. We have to verify a couple of conditions first. So, is the sample random? Well, we can clearly see here that they're going to select a random sample of employees. Therefore, that condition is met.

The second condition is that the distribution of data has to be normal or the sample size has to be greater than or equal to 30. In this case, our sample size for part A, this first part that we're going to tackle over here, is that n is equal to 15. So unfortunately, that's not met. But, is the distribution of data actually normal? Well, again, most of the time you can assume that it is unless you have a lot of evidence to suggest that it's not.

In this case, what's going to happen is some people are going to take some sick days, some maybe take fewer than others. But you should probably assume here that this distribution is going to be normal and not necessarily skewed. So we can actually just go ahead and safely assume that X is normal, and therefore we can proceed with this problem here. So let's take a look at the second step here. The second step is we have to calculate a critical value.

This is where we use our tables or calculators. If we have σ known, we use a Z distribution, but we're not going to use that because we don't have σ. Instead, because it's unknown, we're going to use a t-distribution. All right, let's go ahead and figure out which t-value we're going to use for this. So our α/2, we calculate this as one minus the confidence level, \(0.9 \div 2\). That's going to be 0.05. Now, just be really careful about this because you may already have started to write a number like 1.645. Remember, that's what we used when we used the Z distribution. If the α/2 is 0.05, Z, that critical Z-score is always 1.645.

But because we're working with the t-distribution, which depends on another variable, the degrees of freedom, we can't just immediately write that number down. So we always have to keep track of what our degrees of freedom are in these types of problems. And remember, this depends on the sample size. It's n minus one. So for this problem over here, we have a sample of n = 15 employees.

Therefore, the degrees of freedom is 14. We use both of these numbers together, the α/2 and the degrees of freedom, to figure out what the t-value is, that critical t-score. So this critical t-score of 0.05 is you can go ahead and look this up in a table over here for degrees of freedom of 14. What you're actually going to find here is that this is equal to 1.761. All right.

Now remember, we use this critical t-value because we have to use this in the next equation to calculate the margin of error. The margin of error, remember, depending on whether we're using a Z-distribution or a t, is just going to be slightly different, and we're going to use this formula over here. All right. So, we have that the margin of error is just going to be our t-score, which is going to be our 1.761 times our sample standard deviation, which is 2.4 divided by the square root of the sample size, which in this case was 15. If you work this out, what you're going to find here is that this is 1.09.

That's our margin of error. Okay. So, this is going to be 1.09. Therefore, we can construct our confidence interval, which is the fourth step. Remember, your lower and upper bounds are just going to be, you're just going to have your X̄, your sample mean minus E and then plus E.

All right? So, in other words, we're just going to have our X̄, which is going to be 6.8. So X̄ minus E, X̄ plus E. In other words, this is going to be 6.8, our sample mean minus the 1.09 we just found. Then, we're going to have the 6.8 plus the 1.09 that we just found.

And what you should get over here is you should get that this is 5.71 and this is 7.89. Alright. So, this is our confidence interval. This is at 90% confidence level for this particular problem where we have a sample of 15 employees. All right.

So, in other words, we can be 90% confident that this interval over here actually captures the true average number of sick days taken by all of the employees in the company. Right. That's the population parameter. We're just estimating it. It's going to be somewhere in this interval.

We're 90% confident of that. Okay. Now let's take a look at the second problem over here because almost everything is the same. It says that now suppose that the sample that we got this data from, our X̄ of 6.8, and our sample standard deviation of 2.4. Now let's say that that wasn't 15 employees.

Let's say it was actually 41 employees. Now, how does the 90% confidence interval look now? So basically, how we were sort of interpreting this is that we created our sample, we created our confidence interval using X̄ and E, but now we're just going to go ahead and find our new confidence level using a new sort of sample size of 41 employees.

All right, so let's go ahead and take a look here. We're not going to start all of the steps out from scratch because not really a whole lot is going to change. Right?

You want to purchase one of the new Altima. You randomly select 400 dealerships across the United States and find a mean of $25,000 and sample standard deviation of $2500. Construct and interpret a 94% confidence interval for the true mean price for the new Nissan Altima.

• A. (24996.25, 25003.75); We are 94% confident that the true mean price for the new Nissan Altima falls between $24996.25 and $25003.75.
• B. (24999.25, 25000.24); We are 94% confident that the true mean price for the new Nissan Altima falls between $24999.25 and $25000.24.
• C. (24984.912, 25015.088); We are 94% confident that the true mean price for the new Nissan Altima falls between $24984.912 and $25015.088.
• D. (24764.25, 25235.75); We are 94% confident that the true mean price for the new Nissan Altima falls between $24764.25 and $25235.75.

Answer: (D)

Here’s what students ask on this topic:

How do you construct a confidence interval for a population mean when the population standard deviation is known?

To construct a confidence interval for a population mean when the population standard deviation (σ) is known, follow these steps: First, ensure your sample is random and either normally distributed or has a sample size (n) greater than 30. Use the sample mean (x̄) as a point estimate for the population mean (μ). Calculate the margin of error (E) using the formula: $E=z_{\alpha /2}\sigma /\sqrt{n}$. Here, zα/2 is the critical z-value corresponding to your confidence level. Finally, construct the confidence interval by subtracting and adding the margin of error from the sample mean: $[\mathrm{x̄}-E,\mathrm{x̄}+E]$. This interval estimates the range in which the true population mean lies with a specified level of confidence.

What is the difference between using a z-distribution and a t-distribution for confidence intervals?

The choice between using a z-distribution and a t-distribution for constructing confidence intervals depends on whether the population standard deviation (σ) is known. If σ is known, use the z-distribution, which assumes a normal distribution with a mean of zero and a standard deviation of one. The critical z-value is used to calculate the margin of error. However, if σ is unknown, use the t-distribution, which is similar in shape to the normal distribution but wider and shorter. The t-distribution accounts for additional variability by using the sample standard deviation (s) and degrees of freedom (n-1). As the sample size increases, the t-distribution approaches the normal distribution. The critical t-value is used in place of the z-value, resulting in a larger margin of error, reflecting the increased uncertainty when σ is unknown.

How do you interpret a confidence interval for a population mean?

Interpreting a confidence interval for a population mean involves understanding the range of values within which the true population mean is likely to fall, given a specified level of confidence. For example, if you construct a 90% confidence interval for the mean travel time to work as [55.065, 64.935] minutes, you can say that you are 90% confident that the true mean travel time lies between 55.065 and 64.935 minutes. This does not mean that 90% of the data falls within this range, but rather that if you were to take many samples and construct confidence intervals in the same way, 90% of those intervals would contain the true population mean. The confidence level reflects the degree of certainty in the interval estimate, not the probability of the mean being within the interval.

What are the conditions for using a confidence interval for a population mean?

To use a confidence interval for a population mean, certain conditions must be met: First, the sample must be random, ensuring that each member of the population has an equal chance of being selected. Second, the data should be approximately normally distributed. This can be assumed if the sample size (n) is greater than 30, due to the Central Limit Theorem, which states that the sampling distribution of the sample mean will be approximately normal regardless of the population distribution. If the sample size is less than 30, the data should be checked for normality. Additionally, if the population standard deviation (σ) is unknown, the sample standard deviation (s) should be used, and the t-distribution should be applied instead of the z-distribution.