In a study of proctored and non-proctored tests in an online course, researches
obtained the data below:
Group 1 (Proctored): 𝑛 = 30, 𝑥̅= 74.30, 𝑠 = 12.87
Group 2 (Non-proctored): 𝑛 = 32, 𝑥̅= 80.62, 𝑠 = 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 𝑯𝟎 FAIL TO REJECT 𝑯𝟎
Reason:
e. State a conclusion within the context of the scenario:
Did i do this correctly?
a. Ho:µ1 - µ2 ≥ 0
Ha:µ1 - µ2 < 0
b. test statistic = (74.30-80.62)/√(12.82²/30 + 22.09²/32 = -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.
a. H0: µ1 - µ2 ≥ 0 (There is no significant difference or the mean of non-proctored tests is less than or equal to the mean of proctored tests)
HA: µ1 - µ2 < 0 (There is a significant difference or the mean of non-proctored tests is less than the mean of proctored tests)
b. Test statistic = (74.30 - 80.62) / √((12.87² / 30) + (22.09² / 32)) = -1.387
c. P-value = P(t(60) < -1.387) = 0.085
d. Result: Fail to reject H0
Reason: The p-value of 0.085 is larger than the significance level of 0.05. Therefore, we do not have enough evidence to reject the null hypothesis.
e. Conclusion: Based on the data, we do not have sufficient evidence to conclude that students taking non-proctored tests have a higher mean than those taking proctored tests.
Your work seems to be correct! Here's a breakdown of the steps:
a. The symbolic null and alternative hypotheses are:
H0: µ1 - µ2 ≥ 0 (The mean of students taking non-proctored tests is greater than or equal to the mean of students taking proctored tests)
HA: µ1 - µ2 < 0 (The mean of students taking non-proctored tests is less than the mean of students taking proctored tests)
b. The test statistic is calculated using the formula:
test statistic = (x̄1 - x̄2) / √(s1²/n1 + s2²/n2)
Plugging in the values:
test statistic = (74.30 - 80.62) / √(12.87²/30 + 22.09²/32)
Rounding to 3 decimal places, the test statistic is approximately -1.387.
c. The p-value is the probability of observing a test statistic as extreme as the one calculated (or even further extreme), assuming the null hypothesis is true.
In this case, since the test statistic is negative and we have a one-tailed test to the left, we're interested in the area under the t-distribution curve to the left of the test statistic.
By looking up the test statistic in the t-table with degrees of freedom (df) = 30 + 32 - 2 = 60, we find the p-value to be approximately 0.085.
d. Since the p-value (0.085) is larger than the significance level (0.05), we fail to reject the null hypothesis.
Reason: The p-value represents the probability of observing the results (or more extreme results) given the null hypothesis is true. Since the p-value is not smaller than the significance level, we do not have enough evidence to reject the null hypothesis.
e. Based on the statistical analysis, we do not reject the null hypothesis. This means that we do not have sufficient evidence to conclude that students taking non-proctored tests have a higher mean than those taking proctored tests.