Question

Please compute your z -don't use a package State your conclusion in plain English - not just rejection 1 Students in the online class are suspicious that their schools 60 point loss may have caused their teacher to take out his frustration on the students by giving them a harder than usual exam. They manage to hack into his computer and get the following data: Scores for test Nov 2011 71 74 64 77 58 72 73 79 50 78 43 68 71 59 64 75 72 76 79 52 57 81 73 70 42 64 61 59 75 72 61 63 Scores for test Nov 2010 87 80 62 57 68 34 75 83 89 82 86 83 89 89 87 72 58 30 86 45 82 61 81 29 72 82 85 66 60 29 68 91 86 81 88 82 84 85 86 79 Can they present their case to the dean? Use a z test with an alpha of .05 RR

Answer 1

Z test is used to compare the two population means. The test is performed in following steps,

Step 1: The null and alternative hypotheses are defined as,

0 1t= T11 : OH

HA

Where, $\mu_1$​​​​​​ is the mean test score for Nov 2011 and is the mean test score for Nov 2010.

2

This is a one tailed test (left tailed)

Step 2: The significance level is for the test, . The rejection region is defined by the critical value. The critical value for the z-statistic is obtained from z distribution table for significance level =0.05.

0.05

-1.645 for left tail)

Step 3: The z-statistic is obtained using the formula,

Х2 Х1 z = п2 пi

66.65625 72.975 10.29752 17.95792 32 40

Z-1.873

Step 4: Since,

1.873> -1.645)

The null hypothesis is rejected. Hence we can conclude that there is a significant less mean test score for Nov 2011 compare to Nov 2010.