Question
The MSC was curious to see whether they could say that the mean number of visits per student after COVID is more than during COVID. We randomly took a sample of 64 students that visited the MSC during COVID and the mean number of visits was 2.8 with a standard deviation of 1.05. After COVID we took a random sample of 64 students that visited the MSC and found the mean number of visits per student was 3.2 with a standard deviation of 1.12. Run a hypothesis test at a significance level of 0.10 on the claim that the mean number of visits per student after COVID is greater than during COVID. Answer: 1. Null and Alternative Hypothesis 2. Test Statistic 3. P-Value - just state the p-value 4. Decision 5. Conclusion
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Answer to a math question The MSC was curious to see whether they could say that the mean number of visits per student after COVID is more than during COVID. We randomly took a sample of 64 students that visited the MSC during COVID and the mean number of visits was 2.8 with a standard deviation of 1.05. After COVID we took a random sample of 64 students that visited the MSC and found the mean number of visits per student was 3.2 with a standard deviation of 1.12. Run a hypothesis test at a significance level of 0.10 on the claim that the mean number of visits per student after COVID is greater than during COVID. Answer: 1. Null and Alternative Hypothesis 2. Test Statistic 3. P-Value - just state the p-value 4. Decision 5. Conclusion
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Solution:
1. Null and Alternative Hypothesis:
Null Hypothesis, H0: The mean number of visits per student after COVID is not more than during COVID. (μ <= 2.8)
Alternative Hypothesis, Ha: The mean number of visits per student after COVID is greater than during COVID. (μ > 2.8)
2. Test Statistic:
The test statistic for comparing two means is the t-statistic, given as:
t = (x1 - x2) / sqrt((s1^2/n1) + (s2^2/n2))
where x1 and x2 are the sample means, s1 and s2 are the sample standard deviations, and n1 and n2 are the sample sizes.
Here, x1 = 2.8, x2 = 3.2, s1 = 1.05, s2 = 1.12, n1 = 64, n2 = 64.
t = (2.8 - 3.2) / sqrt((1.05^2/64) + (1.12^2/64))
3. P-Value:
We need to find the p-value for the one-tailed t-test with degrees of freedom (df) = n1 + n2 - 2 = 126.
Using a t-distribution table or calculator, for a significance level of 0.10 and df = 126, the critical t-value is approximately 1.289.
Since the alternative hypothesis is "greater than", we are looking for the p-value in the right tail of the t-distribution.
From the t-distribution table or calculator, the p-value is found to be less than 0.10 (it is usually provided as a range or as "p < 0.10").
4. Decision:
Since the p-value (less than 0.10) is less than the significance level of 0.10, we reject the null hypothesis.
5. Conclusion:
Based on the sample data, at a significance level of 0.10, there is sufficient evidence to support the claim that the mean number of visits per student after COVID is greater than during COVID.
Answer: The p-value is less than 0.10.
1. Null and Alternative Hypothesis:
Null Hypothesis, H0: The mean number of visits per student after COVID is not more than during COVID. (μ <= 2.8)
Alternative Hypothesis, Ha: The mean number of visits per student after COVID is greater than during COVID. (μ > 2.8)
2. Test Statistic:
The test statistic for comparing two means is the t-statistic, given as:
t = (x1 - x2) / sqrt((s1^2/n1) + (s2^2/n2))
where x1 and x2 are the sample means, s1 and s2 are the sample standard deviations, and n1 and n2 are the sample sizes.
Here, x1 = 2.8, x2 = 3.2, s1 = 1.05, s2 = 1.12, n1 = 64, n2 = 64.
t = (2.8 - 3.2) / sqrt((1.05^2/64) + (1.12^2/64))
3. P-Value:
We need to find the p-value for the one-tailed t-test with degrees of freedom (df) = n1 + n2 - 2 = 126.
Using a t-distribution table or calculator, for a significance level of 0.10 and df = 126, the critical t-value is approximately 1.289.
Since the alternative hypothesis is "greater than", we are looking for the p-value in the right tail of the t-distribution.
From the t-distribution table or calculator, the p-value is found to be less than 0.10 (it is usually provided as a range or as "p < 0.10").
4. Decision:
Since the p-value (less than 0.10) is less than the significance level of 0.10, we reject the null hypothesis.
5. Conclusion:
Based on the sample data, at a significance level of 0.10, there is sufficient evidence to support the claim that the mean number of visits per student after COVID is greater than during COVID.
Answer: The p-value is less than 0.10.
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