Question

In: Statistics and Probability

Theres two questions: A professor believes that, for the introductory art history classes at his university,...

Theres two questions:
A professor believes 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. He collects data from a random sample of 250 students in evening classes and finds that they have a mean test score of 80.2. He knows the population standard deviation for the evening classes to be 11.9. A random sample of 300 students from morning classes results in a mean test score of 86.8. He knows the population standard deviation for the morning classes to be 10.2 points. Test his claim with a 95% level of confidence.


A randomly selected group of 35 adult American men was asked if they owned a handgun. 20 said yes. A randomly selected group of 40 American women was asked if they owned a handgun. 21 said yes. Does there appear to be a significant difference between the proportion of adult men and adult women who own handguns?

Solutions

Expert Solution

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 (H​​​​​​0) 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.


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