STAT 200. Guided Exercise 7

Transcription

1 STAT 00 Guided Exercise 7 1. There are two main retirement lans for emloyees, Tax Sheltered Annuity (TSA) and a 401(K). A study in North Carolina investigated whether emloyees with similar incomes differ in their average annual contributions to these lans. The samles are indeendent, random samles, assumed to be normal. Test to see if there is a difference using α.05. This is a small samle difference of means test. The summary statistics from the two samles: TSA 401(K) n 1 15 n 15 Mean 1 $,55 mean $,140 s 1 $ 645 s $ 708 a. Calculate the ooled estimate of the of the two samles (remember to us s ).` ( ) (15 1)645 + (15 1)708 S ( ) S ,644.5 b. Calculate the standard error for this roblem used the ooled estimate of the across the two samles in Part a above sx x s x x c. Conduct the Hyothesis Test to see if there is a difference between the two grous using α.05. Null Hyothesis µ 1 -µ 0 µ 1 -µ 0 Small samle difference of means test; assume normal, ool Test Statistic (z* or t*) t* ( )/47.9 Rejection Region t.05/, 8 d.f..048 Calculation of Test Statistics t*.465 Comarison of Test Statistics with Rejection Region t* < t.05/, 8 d.f.465 <.048 Cannot Reject Ho: µ 1 -µ 0

2 d. Confirm that if the samle sizes are equal for each grou, the ooled is simly the average of the two s. (416,05+501,64)/ 458,644.5 When the two samle sizes are equal, the ooled formula simlifies to an average of the s.. Geneticists have identified EF1 transcrition factor as an imortant comonent of cell roliferation control. The researchers induced DNA synthesis in two batches of serum-starved cells. In one grou of 9 cells (treatment), cells were micro-injected with the EF1 gene. A control grou of 158 cells was not exosed to EF1. After 30 hours, researchers determined the number of altered growth cells in each batch. Test to see if the roortion of growth altered cells for the treatment grou is larger than that of the control. Conduct the hyothesis test using Use α.01. Note: this is a difference of roortions test. For a difference in roortions test where the null hyothesis says the two grous are equal, we should make a ooled estimate of the roortion (using information from both grous) to estimate the standard error of our test. Control EF1 Treated Pooled estimate Total cells Number of growth altered cells for each grou a. The ooled estimate of for this roblem is (denoted as ) and the ooled estimate of q (denoted as q ) : ( ) ( ) & 1 1 b. The standard error for this test is (note the formula to the right):! # σ $ ( ) + 1 q % n1 n " σ ( ) + n1 n.40* q c. Conduct the Hyothesis Test to see if the roortion of growth altered cells for the treatment grou is larger than that of the control. Conduct the hyothesis test using Use α.01. Null Hyothesis P t - P c 0 P t - P c > 0 one-tailed test Large samle difference of roortions test; assume normal, ool Test Statistic (z* or t*) z* ( )/.0547 Rejection Region z Calculation of Test Statistic z* 6.4 Comarison of Test Statistics with Rejection Region z* > z >.33 Reject Ho: P t - P c 0

3 d. What is the -value for the test statistic? A z* 6.4 is very large. We can t even look it u in the standard normal table So just say, <.001 This is very small and good enough for most eole reading your research By the way, the real is ! Females Males Row Total About Right Overweight Underweight Column Totals The following is some data from a survey of college students. The data show the breakdown of resonses to how they feel about their weight by their gender. The resonses to how they erceived their own weight were, about right, overweight, and underweight. The data are given to the right. We can assume that their feelings about their weight is the deendent variable. a. We will focus on the Female vs Male resonse to About Right. I want you to conduct a difference of roortion test to determine if Females and Males had different ercetions about their weight being About Right. Use alha.05. Remember to calculate the ooled roortion for this roblem first, before calculating the standard error. Use alha.05 σ ( ) Null Hyothesis P m - P f 0 q +.680* n1 n ( )/(19+93) (151)/ HYPOTHESIS TEST.0634 P m - P f 0 two-tailed test Large samle difference of roortions test; assume normal, ool Test Statistic (z* or t*) z* ( )/.0634 Rejection Region z Calculation of Test Statistic z*.177 Comarison of Test Statistics with Rejection Region z* < z < 1.96 Cannot reject Ho: P m - P f 0

4 What is the -value of your test? The -value? Look u. in standard normal table Subtract from Multily by for two-tailed test This roblem looks at the salary differences of Male and Female Mid-Level Managers at 0 firms. We will be looking at an Excel file with data on mid-level managers in 0 firms. The salary is given in $1,000s. We want to look at the female samle (n75) and comare it to the male mean level (144) to see if it is lower than that of males (or if males is higher). Use an alha level of.01. Here are the Excel outut: a. Summarize the salary levels for men and women at the comany using the descritive statistics. Male salary slightly larger $144K comared to $140k for females. Mean and median very close for both grous. Variances are very similar for both grous Sread is not too large for either grou they are all managers aroximately 8.8% C.V. Here is the JMP outut for a t-test assuming unequal s. Use it to conduct the test below using an alha level of.01.

5 c. Conduct the Hyothesis Test to see if women earn less than men using α.01. Null Hyothesis µ m -µ f 0 µ m -µ f 0 Large samle difference of means test; assume normal, ool Test Statistic (z* or t*) t*.056 Critical Value at the Rejection Region t.35 Comarison of Test Statistics with Rejection Region The test statistics is.056 it is less that the critical value at the rejection region. We cannot reject Ho: No difference d. What is the -value for this test? All we need do is read the value from the JMP outut. Given this is a two-tailed test we use Prob > t.0415 e. How would you exlain your test result in words. We observed a difference between of the male and female salary of $3,644. When we tested that difference it was not statistically significant at the.01 level for a two-tailed test. This at this level of certainty we cannot be sure if there is a real difference. However, our test would have been significant it the error rate was.05, since the -value for a two-tailed test was f. Could we have assumed the s of the two grous were equal? Brainstorm a bit to decide how we might do this? The easiest way to do this is to divide the two s and see if the result is close to one. In a formal test, this is called an F-test and it follows an F-distribution. When creating this ratio it is best to ut the larger first, so we will look at the rtion of females to males. For our uroses the ratio is: / These s are very close to one another and it would have been reasonable to assume the s were equal. In terms of our test, this would have changed the standard error a small bit, and the degrees of freedom would have increased from to (145-1) + (75-1) 18. However, using a ooled would not have changed the result of the test at alha.01 level.

6 g. I am including the outut assuming equal s from JMP. Did it change the results of the test much? Very little difference whether we assume or not assume equal s the samle sizes are LARGE 4. Earlier we look at some summary data from each state and the District of Columbia for a 0 question test on driving. The score of the test was calculated as 5 oints for each correct question. I administered the same test to 65 graduate and undergraduate students in statistic classes in summer 010. I am going to give you the summary information on the scores for all 65 students as well as a breakdown of the scores for graduate students and undergraduate students from JMP. I want you to do the following: a. Describe the distribution of the test for all 65 students. b. Conduct a two-tailed difference of means test between undergraduate and graduate students. The results are given in JMP. All you need to do is summarize the results. H0: µ undergrad µ grad 0 Ha: µ undergrad µ grad NE 0 two-tailed test Assume equal s. The test statistic from JMP is The -value for a two-tailed test for this value is This result would not meet the criteria for a two-tailed test with alha equal to.05. It would meet the criteria is we were able to accdet a larger Tye I error, for examle at alha equal to.10. Based on an alha level of.05, we would fail to reject the Null Hyothesis and conclude we do not have enough evidence to say the scores of undergrads and graduate students are different.

7 c. Would the results have mattered if we used a one-tailed test that graduate students would score better than undergraduate students? If we used a one-tailed test the -value is.07 that undergraduate students are lower than graduate students (this means the same thing as graduate students score better). Note the JMP software conducted the test as Undergraduate Graduate. Based on this result, we would have been able to reject at alha equal to.05.