Hypothesis Testing (t-Test) - Online Tutor, Practice Problems & Exam Prep

The t-test is a statistical method used to compare the means of populations, particularly when the population standard deviation is unknown. The t-score is calculated using the formula: $t=\frac{\mathrm{x̄}-\mu ₀}{s/\sqrt{n}}$. A larger t-score indicates greater differences between populations. Variations include equal variance, unequal variance, and paired data, each requiring specific calculations to determine significant differences using a t-table.

t-Test

So here we say that the t-test is used to test the mean or average between populations, one of which could be the standard. Now, in order to test the similarities and differences between these populations, we utilize the t-score. To find our t-score, we employ the t-score formula. We use the t-score formula when we don't know what the population standard deviation is. Remember, your standard deviation is just s. As the total number of measurements increases, our standard deviation transitions into our population standard deviation, which is σ. Here, this t-score formula is used when the sample size is less than 30. Size greater than 30 would often utilize software like Microsoft Excel in order to do those large calculations.

Here, our t-score formula is: t = ( x ¯ - μ 0 ) s n , which represents our population average divided by our standard deviation s divided by the square root of the number of samples or measurements, which is n. How does the t-score tell us about similarities and differences between our populations? We're going to say here that the larger the t-score, then the more different the populations are; the smaller the t-score, then the more similar they are to one another.

When we're comparing different populations to one another, we could use three variations of a t-calculated value. Here we can look at our t-calculated for equal variance, for unequal variance, and for paired data. Remember, your variance is just your standard deviation squared. We're going to say here when our standard deviation is equal for both populations, we use these two sets of data. Here, t-calculated will equal the absolute value of the average of population 1 minus the average of population 2 divided by s_pooled times the square root of the measurements divided by the measurements added together. s_pooled would have its own formula here where we're dealing with the standard deviation squared of population 1 times the number of measurements of Population 1 minus 1 plus the standard deviation of Population 2 squared times its number of measurements minus 1. Here on the bottom, we'd have the measurement of Population 1 plus the measurement of Population 2 minus 2. That bottom portion also represents the degrees of freedom involved.

When the variances are not equal between my two populations, then I use these two sets of data. Again, we use 1 to help us figure out what our t-calculated would be. Notice the differences in the formula whether the variances are equal or not. You're not going to be expected to memorize these. Normally, you're given a formula sheet in which you can use them. But always refer to your professor just in case. These calculations, these formulas can get very complicated, so it's always best just to give them to you.

Paired data is used when we have two populations done by completely different methods. Let's say you're trying to analyze the reactivity or fluorescence of some type of material. You have a bunch of different methods that can be used that are very different from one another. In that case, we'd rely on paired data to figure out t-calculated. For the first two, we may be looking at two different populations, but we're using the same type of software and the same types of methods in order to figure out our t-calculated. The whole point of figuring out t-calculated at this point would be then to compare it to our t-table. You'd find your T calculated and you can compare it to your T-table. If your T-calculated happens to be greater than your T-table, remember we looked at our t-table when we're dealing with just figuring out confidence intervals. It'd be that same exact table. If t-calculated were found to be greater than t-table, then you would say there is a significant difference in the average or means of the two populations in which you're examining. If your t-calculated were less, then you would say that there is no significant difference between the average or means between the two populations.

t-Test Calculations

So here it states, a student wishing to calculate the amount of arsenic in cigarettes decides to run 2 separate methods in her analysis. The results shown in parts per million are shown below. Alright. So we have 5 samples for both methods. Here it asks us, is there a significant difference between the 2 analytical methods under a 95% confidence interval? All right. We're dealing with confidence intervals. We know we're going to be dealing with t calculated in some way. Realize here that we're not dealing with just one method with a set of measurements, so we can't just simply use our t score formula to find t calculated and compare it to our t table. Because we're dealing with 2 different methods with their own set of samples, we're going to have to rely on the t test. Now the t test, basically we're going to say here if our t calculated which is what we're going to figure out is greater than our t table, we're going to say that there is a significant difference in terms of the means for these 2 different methods. If we calculate t and we see that it is less than our t table, then we will say there is no significant difference. Now, how do we figure out t calculated? Remember from the previous page that when it comes to our t test, t calculated can be determined in 3 different ways. Either by figuring out we have equal variances and therefore we'd use those set of equations. We have unequal variances, so we'd use a different set of equations. Or if we have paired data in which we use yet another set of equations. Here they're talking about 2 separate methods. So 2 different runs under different situations, therefore this isn't paired data. That means we have to determine if the variances in these two methods, are they equal or are they not equal? Determining that will determine which formula we use for t calculated and for our standard deviations or our degrees of freedom. We have to figure out what their variances are. Remember, variance is just your standard deviation squared. We're going to say here, we first have to figure out what our means are for each. For method 1, we're going to say that our mean or average equals each one of the measurements divided by the total number of measurements divided by 5. So when we do that, that gives us 92.1 for the mean or average of our first method. For the second method, we do the same thing again. Add up each one of the measurements divided by the total number of measurements. So divide by 5. So this equals 92.06. Alright. Now that we've determined that, we're going to now find the standard deviations of the 2. And from that information, we'll be able to determine if they are equal or not equal. That'll determine which set of equations we should use to figure out t calculated. Now we're going to need room guys so let me take myself out of the image. All right. Let's look at method 1. Now remember from previous videos that standard deviation which we're just going to label as s equals square root and we have the summation of each measurement minus the average or mean squared divided by number of measurements minus 1. For method 1, standard deviation equals square root of (110.5 minus the average that we found for method 1 squared, next measurement, 93.1 minus 92.1 squared, 63.0 minus 92.1 squared plus 72.3 minus 92.1 squared, and then finally, 121.6 minus 92.1 squared) divided by n minus 1. n is the number of measurements which is 5 minus 1. Here when we do all that, we get a standard deviation of 24.742. Next, method 2, It's standard deviation, same exact process. So here we have 104.7 minus 92.06 squared plus 95.8 minus 92.06 squared plus 71.2 minus 92.06 squared, plus 69.9 minus 92.06 squared, and then finally, 118.7 minus 92.06 squared, divided by number of measurements which is 5 minus 1. Here when we do that, our standard deviation comes out to 21.27. Remember, your variance is just your standard deviation squared. So it would just be 24.742 squared. So here when we do that, we get a value of 612.167. And then 21.27 squared equals 452.413. Here we have our variances for both. We can see that they are very different values. So we'd say here that the variances are not equal to one another, that tells us which formula to use to figure out my t calculated. We're going to say unequal variances. Variances that are equal to one another are really different by less than 1 from one another. Here, they're different by well over 100 from one another. Alright. So we know that we're dealing with unequal variances. Now we're going to use the formula to figure out t calculated when the variances are not equal. T calculated when the variances are not equal equals the absolute value of (mean of method 1 minus mean of method 2) divided by the square root of (standard deviation 1 squared divided by n1 plus standard deviation 2 squared divided by n2). So that's our formula that we're gonna use to figure out t calculated. All we do now is we input the values that we got. So that's 92.1 minus 92.06 divided by square root of (24.742 squared divided by 5 plus 21.27 squared divided by 5). Now, let's plug in these values here and figure out what number we're gonna get. This right here comes out to be 122.433, and this portion here comes out to be +90.4826 equals. So let's see when we plug that in, what that gives us. So that comes out to being 0.002741 for t calculated. So we have to double check our numbers. That's our t calculated for right now. Now, we are going to have to compare that t calculated to our t table value. But remember, we know that we're dealing with a 95% confidence interval. We know what percentage we're dealing with but we st

t-Test Calculations

In this question, it states that you want to determine if concentrations of hydrocarbons in seawater measured by fluorescence are significantly different than concentrations measured by a second method, specifically based on the use of gas chromatography/flame ionization detection (GCFID). You measure the concentrations of a certified standard reference material, which is 100.0 micromolar. In both methods, you have 7 times. Specifically, you first measure each sample by fluorescence and then measure the same sample by GCFID. The concentrations determined by the two methods are shown below. Alright. So we have these 7 samples being measured by 2 completely different methods. It's paired data. That tells us what formula we need to use in order to figure out our standard deviation as well as our t calculated value. Now here it says calculate the appropriate t statistic to compare the 2 sets of measurements. Again, we're looking at 2 entirely different methods in order to figure out these values. And because they're totally different methods, we're going to compare them by the paired data steps.

We're going to say here, following paired data,

t calculated = | d ̄ | s d ̄ ∗ n

Here, this represents our mean difference in absolute terms divided by our standard deviation times the number of measurements, which is n. Then we're going to say our standard deviation equals:

∑ ( d - d ̄ ) ) 2 n - 1

Now how do we figure out our difference? Well, here, we're going to come up with another column, which is our difference. So what we're going to do here is we're going to take each one of these numbers and subtract them from each other. This is 100.2 - 101.2 or 101.1 which equals - 0.9 . Then we're going to do 100.9 - 100.5 , so that's 0.4 . 99.2 - 100.2 which gives me - 0.3 , 99.9 . We're going to have a 100 and 100.1 - 100.2 , which is - 0.1 , 100.1 - 99.8 which equals 0.3 . 101.1 - 100.7 equals 0.4 and then 100.0 minus 99.9 which equals 0.1 . So, all we've just done is figure out the differences. We are subtracting these values from one another. Next, we're going to calculate what our mean difference is. Just like any mean, we're going to take each one of the differences, add them up together, and divide by the number of measurements. That's going to give me - 0.014 as my mean difference.

Regarding the standard deviation calculation, you take each difference measurement. For instance, 0.9 - 0.014 squared plus 0.4 - 0.014 squared plus etc., until you plug them all into a tedious, but necessary equation. After calculating all these, my standard deviation came out to be 0.47 . So, now we have all we need to figure out what our t calculated is:

t calculated = | 0.014 | 0.47 ∗ 7 equals 0.08 .

Assuming we are dealing with a 95% confidence interval, we find that t calculated, 0.08 , is less than the t table value for 6 degrees of freedom at this confidence level, which is 2.447 . Consequently, there does not appear to be any significant difference between both methods, whether using fluorescence or GCFID. Each method more or less gives us similar means for the two sample populations.

t-Test Calculations

In this question, it says a sample of size \( n \) equals 100 produced the sample mean of 16. Assuming the population deviation is 3, compute a 95% confidence interval for the population mean. All right. We're talking about populations. We're going well above our normal number of measurements within a given population. Remember, we said that if we're going above 30, that usually means that we're not using the t-test but instead the z-test. Now, because I don't give you a z-value here, go to your student's t-table. Look at your students t table and we're dealing with a 95% confidence interval. Look at the column that's dealing with 95%. Since we're dealing with populations that are incredibly large, we're going to look at the degrees of freedom as being equal to infinity. If you line up infinity with your 95% confidence interval, you'll see that your z-score then would be 1.960. That's the logic we use when we're dealing with incredibly large populations like we are in this question. Here, we're going to say that it is our mean or average plus or minus our z-score here times, now we're dealing with our population deviation so that's our population standard deviation. Remember your standard deviation is \( s \) and when we transition to a population standard deviation, it becomes \( \sigma \). It'll be times \( \sigma \) over the number of measurements. Look at the similarities that this has with a typical confidence interval. A typical confidence interval would just be the mean plus or minus your t-score times your standard deviation divided by the square root of \( n \). Again, we transition our standard deviation to the population standard deviation. Because we're dealing with so many numbers that are much greater than normal because we're dealing with a population with a larger dataset, t has transitioned into z. Other than that, we plug in the values and we'll have our answer. Our mean here is 16 plus or minus 1.960 times your deviation which is 3 divided by the square root of 100. Here, when we plug all this into our calculator, it gives us 0.588. So this is 16 plus or minus 0.588. So what does that mean? That means we have \( 16 - 0.588 \) and then we have \( 16 + 0.588 \). So that means that we're 95% confident that our value will lie between 15.412 and 16.588. That would be our level of confidence within this particular question. Remember, when we're going beyond 30, we transition from more of a t-score to a z-score. Here, we're dealing with infinity in terms of degrees of freedom and therefore, because we're dealing with a 95% confidence interval, when we line it up on the t-table, that gives us a score of 1.960. Now that you've seen this one, attempt to do the practice question left here on the bottom. Once you do, come back and see how I approach that same exact practice question.

The average height of the US male is approximately 68 inches. What is the probability of selecting a group of males with average height of 72 inches or greater with a standard deviation of 5 inches?

• A. 84.61 %
• B. 78.81 %
• C. 84.85 %
• D. 79.10 %

Here’s what students ask on this topic:

What is the t-test used for in statistics?

The t-test is a statistical method used to compare the means of two populations, especially when the population standard deviation is unknown. It helps determine if there is a significant difference between the means of the populations. The t-score is calculated using the formula:

$t=\frac{{\overline{x}}_0-\mu )}{\frac{s}{\sqrt{n}}}$

A larger t-score indicates greater differences between the populations. Variations of the t-test include equal variance, unequal variance, and paired data, each requiring specific calculations to determine significant differences using a t-table.

When should you use a t-test instead of a z-test?

You should use a t-test instead of a z-test when the population standard deviation is unknown and the sample size is less than 30. The t-test is more appropriate in these cases because it accounts for the additional uncertainty in the estimate of the population standard deviation. The t-test uses the sample standard deviation (s) and adjusts for smaller sample sizes, making it more reliable for small samples. In contrast, the z-test is used when the population standard deviation is known and the sample size is large (typically n > 30).

How do you interpret the t-score in a t-test?

The t-score in a t-test indicates how much the sample mean deviates from the population mean in units of the standard error. A larger t-score suggests a greater difference between the sample and population means. To interpret the t-score, you compare it to a critical value from the t-table, which depends on the degrees of freedom and the desired confidence level. If the t-score is greater than the critical value, it indicates a significant difference between the means. If it is less, there is no significant difference.

What are the different types of t-tests and when are they used?

There are three main types of t-tests: independent t-test (equal variance), independent t-test (unequal variance), and paired t-test. The independent t-test (equal variance) is used when comparing the means of two independent groups with equal variances. The independent t-test (unequal variance) is used when the variances of the two groups are not equal. The paired t-test is used when comparing means from the same group at different times or under different conditions. Each type of t-test has specific formulas and considerations for calculating the t-score and degrees of freedom.

What is the formula for calculating the t-score in a t-test?

The formula for calculating the t-score in a t-test is:

$t=\frac{{\overline{x}}_0-\mu )}{\frac{s}{\sqrt{n}}}$

where $\overline{x}$ is the sample mean, $\mu$ is the population mean, $s$ is the sample standard deviation, and $n$ is the sample size. This formula is used when the population standard deviation is unknown and the sample size is less than 30.