An instructor hy pothesizes that the mean of the final ex am grades in her statistics class is larger for the male students than that it is for the
female students. The data from the final ex am for the last semester are presented here. Test her claim using = 0.05.
Male Students
36 17 24 39 28
25 38 27 15 21
34 18 22 37 32
Female Students
27 24 33 35 33
25 34 30 29 26
28 23 33 37 25
35 24 23
Find the means first = sum of scores/number of scores
Subtract each of the scores from the mean and square each difference. Find the sum of these squares. Divide that by the number of scores to get variance.
Standard deviation = square root of variance
Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√n
If only one SD is provided, you can use just that to determine SEdiff.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score. Is it less than .05?
I'll let you do the calculations.
To test the instructor's claim, we can perform a hypothesis test. Here's how we can proceed:
Step 1: State the Hypotheses
- Null Hypothesis (H0): The mean final exam grade for male students is equal to or less than the mean final exam grade for female students.
- Alternative Hypothesis (H1): The mean final exam grade for male students is larger than the mean final exam grade for female students.
Step 2: Set the Significance Level (α)
The significance level, denoted by α, determines the probability of rejecting the null hypothesis when it is true. In this case, α is given as 0.05.
Step 3: Calculate the Sample Means
First, let's calculate the sample mean for male students and female students:
Mean (male students) = (36 + 17 + 24 + 39 + 28 + 25 + 38 + 27 + 15 + 21 + 34 + 18 + 22 + 37 + 32) / 15
= 28.6
Mean (female students) = (27 + 24 + 33 + 35 + 33 + 25 + 34 + 30 + 29 + 26 + 28 + 23 + 33 + 37 + 25 + 35 + 24 + 23) / 18
= 29.5
Step 4: Conduct a t-test
We can perform a two-sample t-test to compare the means of the two groups. The formula for the test statistic (t) is:
t = (x̄1 - x̄2) / √[(s1^2 / n1) + (s2^2 / n2)]
where x̄1 and x̄2 are the sample means, s1 and s2 are the sample standard deviations, n1 and n2 are the sample sizes.
Step 5: Calculate the Test Statistic and p-value
Since we are testing if the mean of the male students is larger, we calculate the one-tailed p-value.
Using the values provided:
x̄1 = 28.6 (mean of male students)
x̄2 = 29.5 (mean of female students)
s1 = 7.709 (standard deviation of male students)
s2 = 4.565 (standard deviation of female students)
n1 = 15 (number of male students)
n2 = 18 (number of female students)
Calculating the t-statistic:
t = (28.6 - 29.5) / √[(7.709^2 / 15) + (4.565^2 / 18)]
t = -0.528
To find the p-value, you can use a t-distribution table or a statistical calculator. For a one-tailed test with α = 0.05 and degrees of freedom = 15 + 18 - 2 = 31, the p-value is approximately 0.302.
Step 6: Make a Decision
Compare the p-value to the significance level (α). If the p-value is less than α, reject the null hypothesis. Otherwise, fail to reject the null hypothesis.
Since the p-value (0.302) is greater than the significance level (0.05), we fail to reject the null hypothesis. There is not enough evidence to support the claim that the mean final exam grade for male students is larger than that for female students at a significance level of 0.05.
Note: It's worth mentioning that this analysis assumes that the two samples are independent and that the data is sampled randomly from the populations of male and female students in the instructor's statistics class.