In a study of proctored and non-proctored tests in an online course, researches

obtained the data below:
Group 1 (Proctored): ๐‘› = 30, ๐‘ฅฬ…= 74.30, ๐‘  = 12.87
Group 2 (Non-proctored): ๐‘› = 32, ๐‘ฅฬ…= 80.62, ๐‘  = 22.09
Use of 0.05 significance level to test the claim that students taking non-proctored tests get a
higher mean than those taking proctored tests (use the p-value method). SHOW YOUR WORK.

a. Give the symbolic null & alternative hypotheses. Use correct notation.
H0: HA:
b. Determine the test statistic (round to 3 decimal places): __________________
c. Determine the p-value (round to 3 decimal places):
d. Result (Circle one & justify): REJECT ๐‘ฏ๐ŸŽ FAIL TO REJECT ๐‘ฏ๐ŸŽ
Reason:
e. State a conclusion within the context of the scenario:

Did i do this correctly?

a. Ho:ยต1 - ยต2 โ‰ฅ 0
Ha:ยต1 - ยต2 < 0

b. test statistic = (74.30-80.62)/โˆš(12.82ยฒ/30 + 22.09ยฒ/32 = -1.387

c. 0.085
Df = n1+n2-2 = 60
This is tailed to the left
P-value = P(t(60) < -1.387) = 0.085

d. Fail to reject Ho

Reason: The p value of 0.085 is larger than the significance level of 0.05

e. We do not reject the null hypothesis due to the data not providing enough evidence that the students taking non proctored test get a higher mean than those taking proctored test.

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