SO FAR I HAVE 9.63 FOR THE 1ST 2 COLUMNS
AND 7.56 FOR LAST COLUMN FOR THE VARIANCES. I GOT Z= .0875
SEDIFF =1.2364 SEm=.6019 for 1st set of #s and SEm=1.08 for 2nd set of #s
is that correct and now what?
A total of 23 High School students were admitted to State University. Of those students, 7 were offered athletic scholarships. The school’s guidance counselor looked at their composite ACT scores (given in the table below), wondering if State U might admit people with lower scores if they also were athletes
composite ACT test scores
non atheletes
25
22
19
25
24
25
24
23
21
27
29
26
30
27
26
23
atheletes
22
21
24
27
19
23
17
Test an appropriate hypothesis and state your conclusion. Use alpha = .05 for your test. Show the null and alternative hypothesis, the p-value, your conclusion to reject or not, and finally a summarizing statement regarding your conclusion.
Then create a 90% confidence interval and tell me what it means in terms of the problem
To test the hypothesis and state a conclusion, you will need to perform a hypothesis test using the given data. Here are the steps to follow:
Step 1: Set up the null and alternative hypotheses:
Null hypothesis (H₀): The mean ACT scores for athletes and non-athletes are the same.
Alternative hypothesis (H₁): The mean ACT scores for athletes and non-athletes are different.
Step 2: Calculate the test statistic:
To calculate the test statistic, you will need to find the sample means, variances, and sample sizes for both groups. From the provided data, you already have the variances for both groups.
Let's denote the ACT scores for non-athletes as X₁ and for athletes as X₂.
Mean ACT score for non-athletes (X̄₁) = 9.63
Mean ACT score for athletes (X̄₂) = 7.56
Variance for non-athletes (σ₁²) = 0.0875
Variance for athletes (σ₂²) = 1.2364
Now, calculate the test statistic, Z:
Z = (X̄₁ - X̄₂) / sqrt((σ₁² / n₁) + (σ₂² / n₂))
Here, n₁ and n₂ represent the sample sizes of non-athletes and athletes, respectively.
Step 3: Calculate the standard error and p-value:
To calculate the standard error (SE) and the p-value, you will need the values SE_DIFF and SEm for the two sets of numbers, as provided.
SE_DIFF = 1.08 (for non-athletes)
SEm = 1.08 (for non-athletes)
SEm = 1.2364 (for athletes)
Step 4: Compare the test statistic to the critical value:
With the calculated test statistic, you can compare it to the critical value of the Z-distribution at a significance level of alpha = 0.05. If the test statistic falls within the critical region, you can reject the null hypothesis.
Step 5: Make a conclusion and summarize:
Based on the analysis and the calculated p-value, compare it to the significance level α. If the p-value is less than α, reject the null hypothesis. Otherwise, fail to reject the null hypothesis.
To create a 90% confidence interval, you will use the formula:
CI = (X̄₁ - X̄₂) ± (Z * sqrt(SE₁² + SE₂²))
Here, Z is the critical value of the normal distribution corresponding to a 90% confidence level. The standard errors, SE₁ and SE₂, can be calculated using the provided variances and sample sizes.
The confidence interval will provide a range within which you can be 90% confident that the true difference in mean ACT scores lies.
It's important to note that I will need the sample sizes for both non-athletes and athletes to provide you with the final calculations and interpretations.