A) A market researcher states that she is 95% confident that the average monthly

Question

A) A market researcher states that she is 95% confident that the average monthly sales of a product are between $170,000 and $200,000. What was the mean of the sample she used to arrive at this statement? Explain your answer.

B) Assume that SAT scores vary normally with a standard deviation of 100. How many students should be sampled if we want to estimate the population mean SAT at 99% confidence with a margin of error equal to 25?   [Show your work.]

C) What is the critical t value for a 99% confidence interval using a sample size of 44? Use Excel instead of the table in your text to get a precise answer.

D) A stationery store wants to estimate the mean retail value of greeting cards that it has in its inventory. A random sample of 35 greeting cards indicates an average value of $1.67 and a sample standard deviation of $0.32. Set up a 95% confidence interval estimate of the mean value of all greeting cards in the store’s inventory. (Assume the population is normally distributed.) [HINT: Think about which confidence interval formula you want to use here.] [Show your work.]

E) Given a population with a mean of 70 and a standard deviation of 15, which of the following is most unusual?

An individual with a score of 85?

An individual with a score of 60?

A sample of n = 25 with a mean of 65?

Explain your answer.

Explanation / Answer

SolutionA

confidence interval is

sample mean-margin of error=lower limit

sample mean+margin of error=upper limit

given sample mean-margin of error=170,000

sample mean+margin of error=200,000

add two to get sample mean

2 sample mean=170000+200000

sample mean=185000

SolutionB:

n is sample size

z alpah/2 for 99%=2.576

std deviation=100

n=(2.576*100/25)2

  =106.17

=106(rounding to nearest integer)

SOlutionc:

degrees of feedom=n-1=44-1=43

alpha=0.01

t critical=2.416

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