1. Recall the hypothesis test we considered last time in Class Exercise #6(a)-(f) in Class Handout #3:

2. 6. (a) (b) (c) It is believed that the mean right hand grip strength of men between 20 and 40 years of age in the USA is 86.3 lbs. It is now of interest to perform a hypothesis test concerning the mean grip strength of men between 20 and 40 years of age in the country of Techavia. If we are looking for evidence that the mean grip strength in Techavia is different from 86.3 lbs., state the null and alternative hypotheses for the hypothesis test. H0:H1:H0:H1:  = 86.3 (The mean grip strength is 86.3 lbs.)   86.3 (The mean grip strength is different from 86.3 lbs.) Is the hypothesis test one-sided or two-sided? Describe what it would mean to make a Type I error in this hypothesis test and what it would mean to make a Type II error in this hypothesis test. Since we are looking for evidence that the population mean is different from the hypothesized value 86.3 in either direction, then the test is two-sided Making a Type I error means the mean grip strength is actually 86.3 lbs., but we mistakenly conclude that it is different from 86.3 lbs. Making a Type II error means the mean grip strength is actually different from 86.3 lbs., but we mistakenly conclude that it is equal to 86.3 lbs.

3. (d)Suppose we plan to measure each right hand grip strength in a random sample of 16 men from Techavia. If we assume that either the grip strengths are normally distributed or the sample size 16 is sufficiently large so that the sampling distribution of x is approximately normal, what test statistic would be appropriate for us to use to decide whether to believe H 0 or to believe H 1 ? x – 86.3 s —––  16 If H 0 were true, then would be the t-score for x, where df = and we expect this t-score to be within the bounds of random variation. 15, If H 0 were not true, then we would expect the t-score to be outside the bounds of random variation. Consequently, we can use this t-score as a test statistic to decide whether to believe H 0 or to believe H 1, but we need to choose specific bounds for what should be considered random variation.

4. (e)Find the rejection region for the hypothesis test if (i) a 0.05 significance level were chosen. (ii) a 0.01 significance level were chosen. 1 –  =  — = 2  — = 2 0.95 0.025  = 0.05 t distribution with df = 15 2.131t 0.025 = –2.131– t 0.025 = The rejection region is defined to be all test statistic values t > 2.131 or t < –2.131. 1 –  =  — = 2  — = 2 0.99 0.005  = 0.01 t distribution with df = 15 2.947t 0.005 = –2.947– t 0.005 = The rejection region is defined to be all test statistic values t > 2.947 or t < –2.947. 6.-continued

5. (f)Suppose we actually measure each right hand grip strength in a random sample of 16 men from Techavia, and we find that x = 91.0 lbs. and s = 7.8 lbs. Find the test statistic value, and find the p-value for the hypothesis test. The observed test statistic value is t (or t 15 ) = = = x – 86.3 s —––  16 91.0 – 86.3 7.8 —––  16 2.410 We now need the definition of a p-value. Return to the definitions: p-value (probability value)the probability of obtaining a test statistic value more supportive of H 1 than the test statistic value actually observed, under the assumption H 0 is true

6. (f)Suppose we actually measure each right hand grip strength in a random sample of 16 men from Techavia, and we find that x = 91.0 lbs. and s = 7.8 lbs. Find the test statistic value, and find the p-value for the hypothesis test. The p-value is the probability of obtaining a test statistic value more supportive of H 1 :   86.3 than the test statistic value actually observed, under the assumption H 0 is true. That is, the p-value is the probability that is farther away from zero (0) than the observed test statistic value 2.410. The observed test statistic value is t (or t 15 ) = = = x – 86.3 s —––  16 x – 86.3 s —––  16 91.0 – 86.3 7.8 —––  16 2.410 t distribution with df = 15 2.410– 2.410 From Table 3 of the Statistical Tables, we find that this area must be between 0.01 and 0.025. We now see that the p-value must bebetween 0.02 and 0.05. We denote this by writing 0.02 < p < 0.05.

7. rejection (critical) region a set of test statistic values which lead to rejecting H 0 in favor of H 1 p-value (probability value)the probability of obtaining a test statistic value more supportive of H 1 than the test statistic value actually observed, under the assumption H 0 is true (When we find sufficient evidence against H 0 in support of H 1, we say that we “reject H 0 ” or “accept H 1 ”; when we do not find sufficient evidence against H 0, we say that we “do not reject H 0 ”.) Return to the definition of rejection region for comments on stating the results of a hypothesis test.

8. (g)What should our conclusion in the hypothesis test be, if (i) a 0.05 significance level were chosen? (ii) a 0.01 significance level were chosen? 6.-continued Since the observed test statistic value t (or t 15 ) = 2.410 is in the rejection region corresponding to  = 0.05, 2.131–2.131 2.410 we say that we reject H 0. We can also tell that H 0 should be rejected since p-value < . Results can be written formally as follows: larger than 86.3 lbs. Since the observed test statistic value t (or t 15 ) = 2.410 is not in the rejection region corresponding to  = 0.01, 2.947–2.947 2.410 we say that we do not reject H 0. We can also tell that H 0 should not be rejected since p-value > . Results can be written formally as follows: Since t 15 = 2.410and t 15;0.025 = 2.131,we have sufficient evidence to reject H 0. We conclude that the mean grip strength in Techavia is different from 86.3 lbs. (0.02 < p < 0.05). The results suggest that the mean is

9. statistically significant differencea difference detected by a hypothesis test clinically significant differencea difference which is judged to be large enough to have some practical impact Since t 15 = 2.410 and t 15;0.005 = 2.947, we do not have sufficient evidence to reject H 0. We conclude that the mean grip strength in Techavia is not different from 86.3 lbs. (0.02 < p < 0.05). (h)If we conclude from our hypothesis test that the mean grip strength for Techavia men is significantly different from the mean of 86.3 lbs. for the USA, what practical importance does this have? A hypothesis test is capable only of detecting a statistical significance. In this hypothesis test, the difference between the hypothesized mean of 86.3 lbs. and the sample mean of 91.0 lbs. suggests that the mean for Techavia is almost about 5 lbs. higher than for the USA. Whether or not this difference is of practical significance is a matter of judgment.

10. one-sample t test about a mean The H 0 states a hypothesized value  0 for a population mean . The H 1 is a statement that the hypothesized value  0 is not correct. The test statistic is t (sometimes written t n–1 ) = x –  0 s —–  n Four Steps in a Hypothesis Test Step 1: Step 2: Step 3: Step 4: State the null and alternative hypotheses, and choose a significance level. Collect data, and calculate the value of an appropriate test statistic. Define the rejection region, decide whether or not to reject the null hypothesis, and obtain the p-value of the test. State the results (which should include the observed test statistic value, the tabled value which defines the rejection region, the conclusion, and the p-value), and perform any further analysis which may be required. When the H 0 about a population mean  is rejected, a confidence interval for  can be a follow up analysis to the hypothesis test. ???????????Class Handout #4 summarizes hypothesis tests and confidence intervals concerning one mean, a mean difference, and a difference between two means.

11. Class Handout #4 (Section 1.9, 1.10, material not in text) Definitions one-sample t test about a mean  The H 0 states a hypothesized value  0 for a population mean . The H 1 is a statement that the hypothesized value  0 is not correct. The test statistic is t (sometimes written t n–1 ) = y –  0 s —–  n Statistical Inference Concerning Means (assuming that each random sample is selected from population with a normal distribution or that each sample size is sufficiently large) The data consists of one random sample of n quantitative measurements. We can be (1 –  )100% confident that the population mean  is between y – t  /2 and s ——  n y + t  /2 s ——.  n one sample confidence interval for a mean  The data consists of one random sample of n quantitative measurements.

12. 1. (a) Forbes magazine published data on the best small firms in 1993. (Forbes, November 8, 1993, "America's Best Small Companies,"); these were firms with annual sales of more than $5 million and less than $350 million. The yearly salaries ($1000s) of the chief executive officer (CEO) for the first 20 firms listed are as follows: 145 621 262 208 362 424 339 736 291 58 498 643 390 332 750 368 659 234 396 300 This data is stored in the worksheet CEO_Data of the Excel file M214_Data. A 0.01 significance level is selected to see if there is any evidence that the mean salary for the CEOs is larger than 300 thousand dollars. The firms listed in the FORBES DATA will be treated as a simple random sample of the best small firms. Complete the four steps of the hypothesis test by completing the table titled Hypothesis Test About Mean CEO Salaries.

13. Hypothesis Test About Mean CEO Salaries Step 1H 0 : H 1 :  = Step 2 Step 3 Step 4  = 300  > 300 0.01 (one sided) n = y =s = These statistics can all be obtained by using the Excel spreadsheet named Summary_Statistics, 20 400.8194.503 y –  0 s —–  n t == 400.8 – 300 194.503 ———–  20 =2.318 2.539 t distribution with df = t 0.01 = p-value 0.01 < p < 0.025 do not reject H 0 Since t 19 = 2.318 and t 19;0.01 = 2.539, we do not have sufficient evidence to reject H 0. We conclude that the mean CEO salary is not larger than 300 thousand dollars (0.01 < p < 0.025). 19

14. Step 1H 0 : H 1 :  = Step 2 Step 3 Step 4  = 300  > 300 0.01 (one sided) n = y =s = These statistics can all be obtained by using the Excel spreadsheet named Summary_Statistics, 20 400.8194.503 y –  0 s —–  n t == 400.8 – 300 194.503 ———–  20 =2.318 2.539 t distribution with df = t 0.01 = p-value 0.01 < p < 0.025 do not reject H 0 Since t 19 = 2.318 and t 19;0.05 = 2.539, we do not have sufficient evidence to reject H 0. We conclude that the mean CEO salary is not larger than 300 thousand dollars (0.01 < p < 0.025). 19 (b)Considering the results of the hypothesis test, decide which of the Type I or Type II errors is possible, and describe this error. Since H 0 is not rejected, the Type II error is possible, which is concluding that  = 300 when actually  > 300.

15. (c) (d) 1.-continued Decide whether H 0 would have been rejected or would not have been rejected with each of the following significance levels: (i)  = 0.05, (ii)  = 0.10. H 0 would be rejected with  = 0.05 and with  = 0.10. Use SPSS to do the calculations necessary for the hypothesis test and to create an appropriate graphical display. This data is stored in the SPSS data file ceo. Section E.6 in the appendix of the textbook illustrates how to use SPSS to do the calculations necessary for a one-sample t test about a mean. After selecting the Analyze > Compare Means > One Sample T Test options to display One Sample T Test dialog box, the variable and the hypothesized mean must be entered. Clicking on the Options button allows one to set the confidence level of the confidence interval for the mean displayed by SPSS. When we reject H 0 in a hypothesis test about , a confidence interval can be used to estimate . (In the current hypothesis test, we enter a 99% confidence level, since we used  = 0.01.) A box plot or histogram would be an appropriate graphical display for one sample of quantitative measurements.

16. the sample size, sample mean, and sample standard deviation the estimated standard error of the mean s —–  n the t statistic and degrees of freedom The p-value displayed on the SPSS output is for a two sided test; this must be divided by 2 when doing a one sided test. Consequently, the exact p-value for the hypothesis test in part (a) is 0.032/2 = 0.016. hypothesized mean

17. 2.Measurements of body temperature (BT) in degrees Fahrenheit and heart rate (HR) in beats per minute were derived from a data set presented in Mackowiak, P. A., Wasserman, S. S., and Levine, M. M. (1992), "A Critical Appraisal of 98.6 Degrees F, the Upper Limit of the Normal Body Temperature, and Other Legacies of Carl Reinhold August Wunderlich," Journal of the American Medical Association, 268, 1578-1580. The resulting data is as follows: Males BT 96.3 96.7 96.9 97.0 97.1 97.1 97.1 97.2 97.3 97.4 97.4 HR 70 71 74 80 73 75 82 64 69 70 68 BT 97.4 97.4 97.5 97.5 97.6 97.6 97.6 97.7 97.8 97.8 97.8 HR 72 78 70 75 74 69 73 77 58 73 65 BT 97.8 97.9 97.9 98.0 98.0 98.0 98.0 98.0 98.0 98.1 98.1 HR 74 76 72 78 71 74 67 64 78 73 67 BT 98.2 98.2 98.2 98.2 98.3 98.3 98.4 98.4 98.4 98.4 98.5 HR 66 64 71 72 86 72 68 70 82 84 68 BT 98.5 98.6 98.6 98.6 98.6 98.6 98.6 98.7 98.7 98.8 98.8 HR 71 77 78 83 66 70 82 73 78 78 81 BT 98.8 98.9 99.0 99.0 99.0 99.1 99.2 99.3 99.4 99.5 HR 78 80 75 79 81 71 83 63 70 75

18. Females BT 96.4 96.7 96.8 97.2 97.2 97.4 97.6 97.7 97.7 97.8 97.8 HR 69 62 75 66 68 57 61 84 61 77 62 BT 97.8 97.9 97.9 97.9 98.0 98.0 98.0 98.0 98.0 98.1 98.2 HR 71 68 69 79 76 87 78 73 89 81 73 BT 98.2 98.2 98.2 98.2 98.2 98.3 98.3 98.3 98.4 98.4 98.4 HR 64 65 73 69 57 79 78 80 79 81 73 BT 98.4 98.4 98.5 98.6 98.6 98.6 98.6 98.7 98.7 98.7 98.7 HR 74 84 83 82 85 86 77 72 79 59 64 BT 98.7 98.7 98.8 98.8 98.8 98.8 98.8 98.8 98.8 98.9 99.0 HR 65 82 64 70 83 89 69 73 84 76 79 BT 99.0 99.1 99.1 99.2 99.2 99.3 99.4 99.9 100.0 100.8 HR 81 80 74 77 66 68 77 79 78 77 A 0.10 significance level is selected to see if there is any evidence that the mean heart rate for males is different from 72 beats per minute.

19. (a) 2.-continued The 65 males in the data set will be treated as a random sample. Use SPSS to do the calculations necessary for the hypothesis test and to create an appropriate graphical display. Then, complete the four steps of the hypothesis test by completing the table titled Hypothesis Test About Mean Heart Rate of Males. The data is stored in the SPSS data file metabolism. Before using the Analyze > Compare Means > One Sample T Test options in SPSS, we must first select only the males in that data set as follows: Select the Data> Select Cases options to display the Select Cases dialogue box, and select the If condition is satisfied option. Click on the If button to display the Select Cases If dialogue box. From the list of variables on the left, select the variable sex, and click on the arrow button pointing to the right. Either by use of the buttons in the dialog box or by direct typing, finish the formula so that it reads sex = 1. Click on the Continue button, and click on the OK button, after which you will now notice that a new variable has been added to indicate which cases are to be included and which are to be excluded. In case we reject H 0 and want to estimate the mean with a confidence interval, set the confidence level in SPSS to be 90%, since we have  = 0.10.

20. A box plot or histogram would be an appropriate graphical display for one sample of quantitative measurements.

21. Hypothesis Test About Mean Heart Rate of Males Step 1H 0 : H 1 :  = Step 2 Step 3 Step 4  = 72   72 0.10 (two sided) n = y =s = These statistics can all be obtained from the SPSS output. 65 73.375.875 t =1.879 1.671 t distribution with df = t 0.05 = p-value 0.05 < p < 0.10 reject H 0 Since t 64 = 1.879 and t 64;0.05 = 1.671, we have sufficient evidence to reject H 0. We conclude that the mean heart rate for males is different from 72 beats per minute (0.05 < p < 0.10). The data suggest that the mean heart rate for males is larger than 72 beats per minute. –1.671 from the Student’s t distribution table from the SPSS outputp = 0.065 or (P = 0.065) 64

22. (b) (c) (d) (e) 2.-continued Considering the results of the hypothesis test, decide which of the Type I or Type II errors is possible, and describe this error. Decide whether H 0 would have been rejected or would not have been rejected with each of the following significance levels: (i)  = 0.05, (ii)  = 0.01. For the next class, see if you can finish this exercise by answering parts (b) to (e). Considering the results of the hypothesis test, explain why a 90% confidence interval for the mean heart rate for males would be of interest. Then find and interpret the confidence interval. Does the difference between the sample mean heart rate and the hypothesized mean heart rate represent a clinically significant difference? Why or why not?

23. Before submitting Homework #4, check some of the answers (if you haven’t done so already) from the link on the course schedule: http://srv2.lycoming.edu/~sprgene/M214/Schedule214.htm