Question

Suppose we measure how much sleep students in first and second year at UVic get per night. A sample of 21 first year students slept an average of 7.9 hours per night with a standard deviation of 1.3 hours. A sample of 16 second year students slept an average of 6.8 hours per night with a standard deviation of 2.1 hours. Suppose we want to test if first year students sleep more than second year students. Assuming the number of hours slept is normally distributed for both first year students and second year students, what is the estimated standard error of the random variable that should be used for our test? Give your answer rounded to 4 decimal places. Answer:

Answer 1

1. Let $\mu_1,\mu_2$ denote number of hours slept for first year students and second year students respectively.

Hypotheses : To test null hypothesis Ho Mi - U2 = 0 against alternative hypothesis H 41-42 > 0 This is a one tailed right sided test as the alternative hypothesis contains >' sign. Here, 1 = 7.90, 81 = 1.30, n1 = 21,02 = 6.80, S2 = 2.10, n2 = 16 The test statistic can be written as: (21-22-0 t= + VM 122 which under Ho follows at distribution with 23 df. where + + 2.102 )2 16 ni df = 1.302 21 (1.3022 21 21-1 23 .)2 + + 12-1 1-1 16-1 and standard error for the difference between means sz S SE= + 1.302 21 + 2.102 16 0.596742 ni n2 Decision rule / Rejection region : We reject H, at 0.01 level of significance if P-value < 0.01 or if tstat > to.01,df 0.4- >0.3 20.2- 0.1- 0.0 2 t Now, Value of the test statistic: The value of the test statistic is (7.90 – 6.80) tstat = 1.843342 1.843 V 1.30 21 + 2.102 16 associated degrees of freedom = 23

2.

Let us denote the difference

d = length of symptoms before medication - length of symptoms after medication

Let ud denotes the population mean of difference. Hypotheses : To test null hypothesis Hd = 0 against alternative hypothesis H1 :pid < 0 This is a one sided left tailed test as alternative hypothesis H1 contains only < sign. Let à denotes the sample mean of difference, sd denotes the sample standard deviation of difference and n denotes the sample size. Here, d = 11.0000, så= 369.6000, sd = 19.2250, n=6 The test statistic can be written as: vnd – 0.0) te sd which under H, follows at distribution with n - 1 df. Decision rule / Rejection region : We reject H, at 0.05 level of significance if P-value < 0.05 or if tobs <- 0.05,n-1 =-2.015048 0.3 2.0.2 0.1 0.0 Now, Value of test statistic: The value of the test statistic tobs = V6(11.0000 - 0) 19.2250 = 1.401530 ~ 1.40 associated degrees of freedom = n - 1 = 5 The critical value = critical = -10.05,5 = -2.015048 = -2.015 P-value = P(tn-1 < tobs) = P(t; <1.401530) = 0.8900132 0.8900 Conclusion : Since p-value > 0.05 and tobs * tcritical = -2.015048, so we fail to reject H, at 0.05 level of significance Hence, we can conclude that mean difference is not significantly less than 0.