Question

Grades on a standardized test are known to have a mean of 1000 for students in the US. The test is administered to 453 randomly selected students from Queens College, and they obtained an average grade of 1013 and a standard deviation of 108. a. Construct a 95 % confidence interval for the true average test score for Queens College students. b. With 5% significance level, Is there a statistical evidence that Queens College students perform differently than other students in the US? The same 453 students selected earlier are now given a two-hour tutoring session and then asked to take the test a second time. The average change in their test scores is 9 points, and the standard deviation of the change is 60 points. Assume the changes are from a Normal distribution. c. You are asked by the school administration to study whether students have performed better in their second attempt. State in terms of , the relevant null and alternative hypothesis in conducting this study. d. Compute the t statistic for testing against ; obtain the p- value for the test. e. Do you reject at the 5% level? At the 1% level? f. Provide a short summary of your conclusions from this study. Comment on the practical versus statistical significance of this estimate.

Answer 1

Answer:

a:  1003 < µ < 1023

b:  See attached photo for answers

c:  see below

d and e:  See second attached photo

f:  See below

Step-by-step explanation:

a.  to find a confidence interval, first find the error with the formula,

E = Z(σ/√n)  where Z is the value of the z-score that relates to the confidence level, σ is the standard deviation of the population (or of the sample if the population standard deviation isn't given), and n is the sample size

For this problem, E = 1.96(108/√453) = 9.95 = 10

Now add and subtract that value form the sample average...

x - E < µ < x + E      becomes    1013 - 10 < µ 1013 + 10

simplifies to 1003 < µ < 1023

b.  See attached photo for hypothesis test results...

c.  H0:  µ = 1013  

    Ha:  µ > 1013 (claim)

They want to test if it's greater than their first attempt,  so this claim becomes the alternate hypothesis.  We're testing against their first attempt, that's why µ = 1013, not 1000

d and e.  See second attached photo for hypothesis test results...

f.  There is enough evidence to support the claim that the students second attempt was an improvement from their first attempt.  They happened to take a 2 hour tutoring course between tests.  This doesn't prove that the tutoring was effective thought.  It very well might have been effective, but also, the students taking the test again means they've already seen the test and know what to expect, which could have also boosted their scores.