T-test for the Mean of a Population: Unknown population standard deviation

# T-test for the Mean of a Population: Unknown population standard deviation

## T-test for the Mean of a Population: Unknown population standard deviation

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##### Presentation Transcript

1. T-test for the Mean of a Population: Unknown population standard deviation Here we will focus on two methods of hypothesis testing: the critical value approach and the p-value approach.

2. We saw in the standard deviation in the population known case that when we do not know the true value of the population mean for a quantitative variable an hypothesis test can be carried out utilizing the z calculation (x bar minus mu under Ho:)/standard error of the mean. When the population standard deviation, sigma, of the variable is unknown we have to rely on the t distribution. Plus in the calculation of the standard error we will use the sample standard deviation. The t statistic = (x bar minus mu under Ho:)/ standard error of the mean. Let’s work a few problems.

3. Example For a company when they look at the past the have seen the average dollar amount on an invoice be \$120. Over time this will be monitored and they will see if this changes. The question now is about whether or not the population mean is still 120. We will make this the null hypothesis. So we have Ho: μ = 120 and H1: μ≠ 120 a) With the critical value approach the value of alpha has to be determined and say we have alpha = .05. When the alternative hypothesis has a not equal to sign we have a two tail test. This means we have .025 in each tail. But since we do not know the population standard deviation we have to use the t distribution. With a sample size of 12 we look in the df = n – 1 = 12 – 1 = 11 row. The critical t’s are thus 2.2010 and -2.2010. Let’s see what this looks like in a graph on the next slide

4. .025 alpha/2 = .025 lower Critical t = -2.2010 Upper Critical t = 2.2010 Let’s review what we have done. We have a null and alternative hypothesis. We have an alpha value and a sample size we will use. The critical values of t break up the t distribution into rejection and acceptance of the null hypothesis regions. Our decision rule will be this: If when we take a sample and calculate both a sample mean and the associated t value, called the t test statistic (and I will write tstat), if the tstat is less than the lower critical value or greater than the upper critical value we will reject the null. If the tstat is in the middle of the critical values we do not reject the null.

5. Now say we get an actual sample of 12 invoices and we see the sample mean is 112.85 and the sample standard deviation is 20.80. The tstat from the sample is (112.85 – 120)/(20.804/sqrt(12)) = - 1.19. Since the value of the tstat is -1.19 and since this value is in the middle of the critical values we do not reject the null. b) To proceed with the p-value approach to hypothesis testing I would like us to explore the t distribution with df = 11 row. Let’s see this on the next slide.

6. T distribution with DF = 11 .25 is area under curve .10 is area under curve .05 is area under curve .025 is area under curve .01 is area under curve .005 is area under curve -3.1058 -2.7181 -2.2010 -1.7959 -1.3634 -0.6974 .0.6974 1.3634 1.7959 2.2010 2.7181 3.1058 Here I have marked off on the t distribution the positive and negative values. On the next slide I reproduce this with the tails colored in for when alpha is picked to be .05.

7. T distribution with DF = 11 .25 is area under curve .10 is area under curve .05 is area under curve .025 is area under curve .01 is area under curve .005 is area under curve -3.1058 -2.7181 -2.2010 -1.7959 -1.3634 -0.6974 .0.6974 1.3634 1.7959 2.2010 2.7181 3.1058 So, with alpha = .05 the critical t’s are -2.2010 and 2.2010. Next we take the sample mean and calculate the tstat. Again, in our example we had -1.19. The -1.19 occurs here on the number line. This falls between -1.3634 and -.06974. The tail areas for these two values are 0.10 and 0.25, respectively. 7

8. The tail area for -1.19 is thus between 0.10 and 0.25. This is the basis for the p-value. But, because of the way the t distribution shows up in our book the best we can say about the tail area for the tstat is between 0.10 and 0.25. Since our alternative hypothesis H1 is a not equal to sign we have to double the tail area for -1.19 and so we say the p-value is between 0.20 and 0.50. (A computer or better table would have us see the tail area doubled would be .259 – we do not need that here.) Here is how we use the p-value approach. If the p-value is less than or equal to alpha reject the null, otherwise do not reject the null. In our example the p-value is at least 0.20 which is > .05 so we do not reject the null.

9. Problem 62a pages 296, 297 What is viscosity? Is it like elasticity? From the problem we see Ho: μ = 15.5 and H1: μ≠ 15.5. With alpha = .10 and a two tail test we put .05 in each tail and see with df = 119 the critical values are -1.6577 and 1.6577 (hey Parker, dude, these are values with df=120 – you are correct, but our book does not have a df = 119 so go to the closest df we do have). We see this set up on the next slide. Now, if from the sample we get a tstat less than -1.6577 or greater than 1.6577 we will reject the null.

10. alpha = .10/2 Critical t = 1.6577 -1.6577 To get our tstat we have to use the data given and find the sample mean and sample standard deviation. Using the computer, I find the mean to be 14.9775 and the standard deviation to be 1.01. So, the tstat is (14.9775 – 15.5)/(1.01/sqrt(120)) = -5.67 This is to the left of the lower critical value so we can reject the null. b. Notice in the t table that in the df = 120 row (the closest df we have) the value 2.6174. On the negative side we would have -2.6174. The tail area is .005. Our tstat is even farther out and thus has a tail area less than .005 and when we double this we have a p-value less than .01 Since the p-value < alpha (0.10 here) we reject the null.