QUESTION 1 A major department store chain is interested in estimating the average amount its credit card customers spent on their first visit to the...

QUESTION 1

A major department store chain is interested in estimating the average amount its credit card customers spent on their first visit to the chain's new store in the mall. Fifteen credit card accounts were randomly sampled from the normally distributed population and analyzed with the following results:  = $50.50.  The population variance is 400. A 95% confidence interval for the average amount the credit card customers spent on their first visit to the chain's new store in the mall is:

1. $50.50 ± $9.09.
2. $50.50 ± $10.12.
3. $50.50 ± $11.08.
4. None of these choices.

A survey of an urban university showed that 750 of 1100 students sampled attended a home football game during the season. Using the 99% level of confidence, what is the confidence interval for the proportion of students attending a football game?

[0.767, 0.814]

[0.6550, 0.7050]

[0.6456, 0.7179

[0.6795, 0.6805]

The 95% Confidence Interval for the weights (in pounds) of a special type of rock has found to be 4.3, 8.46. What will be the margin of error? (Hint: Sample mean +/- margin of error = confidence interval)

In a large city, the average number of lawn mowings during summer is normally distributed with mean μ and standard deviation σ = 9.3. If I want the margin of error for a 90% confidence interval to be ±3, I should select a simple random sample of size (4 decimal points) 

In a large city, the average number of lawn mowings during summer is normally distributed with mean μ and standard deviation σ = 11.5. If I want the margin of error for a 99% confidence interval to be ±4, I should select a simple random sample of size (4 decimal points) 

The standard error of the mean for a sample of size 132 is 25. In order to cut the standard error of the mean in half (to 12.5) we must select a sample size of  ...?

21 credit card holders are selected at random. For each, their current credit card balance is recorded. The average for these 21 people is = $600. Assume that the current balance of all credit card holders follows a normal distribution with unknown mean μ, and that a 85.3% confidence interval for μ is found to be $600 ± 34. Find the standard deviation σ of the population?