Question

In: Statistics and Probability

Continuing the discussion about SAT scores, these parameters are for all people who took the SAT,...

Continuing the discussion about SAT scores, these parameters are for all people who took the SAT, not just students who went to college. However, there are some assumptions that people with higher SAT scores will make certain choices. One such choice is the decision to go to college. We do not know the data (mean and standard deviation) for students who go to college, but we can estimate it with a sample. We took a random sample of 27 first year students. Results of these tests are summarized below. ? = 27 ? = 1170 ? = 143 a. Calculate the test statistic (the thing that you will use to look up in a table) using the hypotheses: ??: ?????????? = 1086, the average SAT score of first year students is 1086 (average) ??: ?????????? > 1086, the average SAT score of first year students is greater than 1086 b. State a conclusion using either a p-value or a critical value using a sentence. c. Regardless if you rejected the null hypothesis or not, use the data to find the bounds for a 95% confidence interval to estimate the average score of the students in the current first year class. Your answer does not need to be a sentence

Solutions

Expert Solution

(1) Test for a single independent sample, right tailed test., standard deviation unknown and n < 30

We use a students t distribution

The test statistic is given by the equation:

(b) The p value for df = 26, right tailed = 0.0026

The critical value, right tailed at = 0.05 is 1.71

The Decision Rule:

If p value is < , Then Reject H0.

If t observed is > t critical, the Reject H0.

The Decision:

Since p value is < 0.05, Reject H0

Since t observed (3.05) is > 1.71, Reject H0.

The Conclusion: Reject H0. There is sufficient evidence at the 95% significance level that the average SAT scores of students who go to college is greater than 1086.

________________________

The 95% Confidence level

The Confidence Interval is given by ME

ME = t critical * s / Sqrt(n) = 56.58

The Lower Limit = - ME = 1170 - 56.58 = 1113.42

The Upper Limit = + ME = 1170 - 56.58 = 1226.58

The 95% Confidence interval is (1113.42, 1226.58)

___________________________________________

orchestra answered 2 years ago
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