• A

• B

• C

• D

• E

• A

• B

• C

• D

• E

Mr. Reinecke wants to estimate the proportion of seniors who will go to college next year with 96% confidence and a margin of error of no more than 5%. What size sample does he need to take?

• 422

• 419

• 410

• 400

Mrs. Soto polled 1000 adults and estimated with 90% confidence that the proportion of adults who have dessert after dinner is .75. Ms. Libner thinks that 90% confidence isn't enough and wants to do a 95% confidence. How would a margin of error for 95% compare with a ME for 90% if the same sample size was used?

• It would be smaller because a higher confidence means a smaller MOE

• It would be larger because a higher confidence means a larger MOE

• It would be the same because the sample size is the same

• Won't be able to tell without knowing the true parameter

You want to calculate a 98% confidence interval for a population mean with a sample of n = 20. What is the appropriate t*?

• 2.326

• 2.528

• 2.539

• 2.518

The heights of 5-year old boys are approximately normal. The mean height of a random sample of 37 5-year old boys is found to be 30 inches with a standard deviation of 3 inches. What is the standard error of x̄?

• .493

• 1.32

• 5.77

• 1.96

Mrs. Soto wants to find the sample size needed to find with 95% confidence the mean amount of time her stats students spent on homework during the snow days. The population is approximately normal and the known standard deviation is 6 minutes. How large should the sample size be if she wants a margin of error of +- .5 minutes?

• 553

• 551

• 600

• 601

Scores on the mathematics part of the SAT exam in a recent year were roughly Normal with mean 515 and standard deviation 114. You choose an SRS of 100 students and average their SAT Math scores. Suppose that you do this many, many times. Which of the following are the mean and standard deviation of the sampling distribution of x̄ ?

• Mean = 515; SD = 114

• Mean = 515; SD = 114/√100

• Mean = 515/100; SD = 114/√100

• Mean = 515/100; SD = 114/100

• Mean = 515/100; SD = 114/√100

Why is it important to check the 10% condition before calculating probabilities involving x̄?

• To reduce the variability of the sampling distribution of x̄

• To ensure that the distribution of x̄ is approximately Normal

• To ensure that we can generalize the results to a larger population

• To ensure that x̄ will be an unbiased estimator of μ.

• To ensure that the observations in the sample are close to independent.

One reason for using a t distribution instead of the standard Normal curve to find critical values when calculating a level C confidence interval for a population mean is that

• z can be used only for large samples

• z requires that you know the population standard deviation

• z requires that you can regard your data as an SRS from the population

• z requires that the sample size is at most 10% of the population size

• a z critical value will lead to a wider interval than a t critical value