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.

Yes, you have done the calculations correctly.

a. The symbolic null and alternative hypotheses are:
Ho: ยต1 - ยต2 โ‰ฅ 0 (The mean of Group 1 is greater than or equal to the mean of Group 2)
Ha: ยต1 - ยต2 < 0 (The mean of Group 1 is less than the mean of Group 2)

b. To determine the test statistic, you can use the formula for the difference in means:
test statistic = (xฬ„1 - xฬ„2) / โˆš((s1^2 / n1) + (s2^2 / n2))
Plugging in the given values:
test statistic = (74.30 - 80.62) / โˆš((12.87^2 / 30) + (22.09^2 / 32)) = -1.387 (rounded to 3 decimal places)

c. To determine the p-value, you need to find the probability of obtaining a test statistic value more extreme than the observed test statistic value under the null hypothesis. Since this is a one-tailed test in the left direction, you need to find the area to the left of the observed test statistic value in the t-distribution. The degrees of freedom (df) can be calculated using the formula: df = n1 + n2 - 2 = 30 + 32 - 2 = 60.
P-value = P(t(df) < -1.387) = 0.085 (rounded to 3 decimal places)

d. Since the p-value is larger than the significance level of 0.05, we fail to reject the null hypothesis.

e. Based on the results of the test, we do not have enough evidence to conclude that students taking non-proctored tests get a higher mean than those taking proctored tests.