In: Statistics and Probability
Solution :
1) The null and alternative hypotheses are as follows :
Where, are the mean test scores of the students in the morning and evening classes respectively, for the introductory art history classes at the University.
To test the hypothesis we shall use z-test for testing the equality of two population means. The test statistic is given as follows :
Where, are sample means for morning and evening classes respectively, is the hypothesized value of difference in the mean scores, are the population standard deviations for morning and evening classes respectively and are sample sizes for morning and evening classes respectively.
We have,
The value of the test statistic is 1.6743.
Since, our test is right-tailed test, therefore we shall obtain right-tailed p-value for the test statistic. The right tailed p-value is given as follows:
p-value = P(Z > value of the test statistic)
p-value = P(Z > 1.6743)
p-value = 0.0470
The p-value is 0.0470.
Confidence level = 95% = 0.95
Hence, significance level = 1 - 0.95 = 0.05
(0.0470 < 0.05)
Since, p-value is less than the significance level of 0.05, therefore we shall reject the null hypothesis (H0) at 0.05 significance level.
Conclusion : At 95% confidence level, we have enough evidence to support the professor's claim that, for the introductory art history classes at his university, the mean test score of students in the evening classes is more than 5 points lower than the mean test score of students in the morning classes.