Suppose 400 randomly selected alumni from Miami University were asked to rate the university’s counseling

Question 1: (10 points)

Suppose 400 randomly selected alumni from Miami University were asked to rate the university’s counseling services on a 1 to 10 scale.  The sample mean was found to be 8.6.  Assume the population standard deviation is known to be 2.0.

a.    John Smith computes the 95% confidence interval for the average satisfaction score as              8.6 ± 1.96(2). What is his mistake?

b.    After correcting his mistake in part (a), he states “I am 95% confident that the sample mean falls between 8.404 and 8.796”.  What is wrong with his statement?

c.    Finally, in his defense of using the Normal Distribution to determine the confidence coefficient he says, “Because the sample size is quite large, the population of alumni rating will be approximately Normal.”  Explain to John his misunderstanding and correct his statement.

Question 2:(10 points)

The mean score for all U.S. high school seniors taking the SAT college entrance exam equals 500. A study is conducted to see whether a different mean applies to Canadian seniors. For a random sample of 100 Canadian seniors, suppose the mean and standard deviation on this exam equal 508 and 100.

a.       Set up hypotheses for a significance test, and compute the test statistic.

Here we want to see whether a different mean applies to Canadian seniors than U.S seniors or not so the hypotheses are,

Test statistic =

b.      The P-value is 0.43. Interpret it, and make a decision about H₀, using a significance level of 0.05.

The p-value is the probability of observing the value of the test statistic or extreme under the null hypothesis. So in this case if the null hypothesis was true then observing the value of the test statistic 0.8 or extreme (in both sides as it’s a two tail test) has probability 0.43. As the p-value is larger than the significance level of 0.05 so we are not rejecting the null hypothesis and the conclusion is the mean is not significantly different than hypothesized mean 500.

c.       If the decision in (b) was in error, what type of error is it?

The decision in part b is not to reject the null hypothesis so if its in error that means we are not rejecting the null hypothesis though it is false so it’s a Type II error.

Question 3 (10 points)

A nationwide study of American homeownersrevealed that 65% have one or more lawnmowers. A lawn equipment manufacturer located in Omaha, feels the estimate is too lowfor households in Omaha and predicts this amount to be higher. Can the value 0.65be rejected if a survey of 490 homes in Omahayields 336 with one or more lawn mowers? Useα = 0.05. (Show all steps)

Here we want to see whether the percentage of American homeowners who have one or more lawn mowers is more than 65% or not so the hypotheses are,

From the alternative hypothesis we can see that it’s a one tail (right tail) test thus for the significance level of 5% or 0.05 the critical region is,

Test statistic >Critical value

Now here we are testing for one sample proportion so we should use a Z test.

Test statistic =

Critical value = Z(0.05) = 1.645

As test statistic > Critical value i.e. the test statistic is falling in the rejection region so we are rejecting the null hypothesis based on the given information and concluding that the data shows evidence that the true percentage is actually more than 65%.

Question 4: (10 points)

The ages (in years) at first transplant for a sample of 11 heart transplant patients are as follows:

54 42 51 54 49 56 33 58 54 64 49. The summaries for these data are: n = 11; Y bar = 51.27; and s = 8.26;

a.    Calculate a 95% Confidence Interval for the mean age at first transplant

Here the population standard deviation is unknown as well as the sample size is small so we should use a t-multiplier thus the 95% confidence interval is,

95% CI =

b.    Additionally, calculate the p-value for the Hypothesis Test Ho: μ = 50 vs. Ha μ ≠ 50 and test this against α = .10

We can see that the test is a two tailed test.

Test statistic =

P-value =