MTH93 Unit 5 Review

what is the positive critical value if you are 88 percent confident

construct a 99% confidence interval for the following:

24 people were randomly selected to take part in a pizza survey on if they prefer thin or stuffed crust on their pizza. according to a recent (similar) survey it was found on average 62% of people preferred thin crust and the standard deviation was 5%. assume this data is normally distributed and round your answers off to the hundredths!

A marketing specialist for Naples pizza wants to check on average how many people tend to pick thin crust over stuffed crust. He wants to be 96% confident and accurate within 1.5% of the mean. From a previous survey he found the standard deviation was 5%, how many people would need to be surveyed for this?

Create a confidence interval for the following given information:

Mean = 32

n=17

sample standard deviation= 3.25

you want to be 90% confident and this distribution follows the normal distribution

round your answers off to the hundredths!

Answer explanation

is the T distribution!

Would you use the standard normal table or t-distribution table to create a confidence interval for the following information:

Mean=18, standard deviation= 2.5, n=20, this follows a normal distribution.

Answer explanation

z table! n<30 but its normal and is not the sample standard dev!

Would you use the standard normal table or t-distribution table to create a confidence interval for the following information:

Mean=211, sample standard deviation= 10, n=23, this follows a normal distribution.

Answer explanation

n<30 and the sample standard dev is given!

Would you use the standard normal table or t-distribution table to create a confidence interval for the following information:

Mean=75, standard deviation= 14, n=60, this doesn't follows a normal distribution.

Answer explanation

n>30, doesnt need to follow a normal distribution since n is so big!

what is the negative critical value if you are 78 percent confident?

what is the positive critical value if you are 62 percent confident?

A marketing specialist for Naples pizza wants to check on average how many people tend to pick thin crust over stuffed crust. He wants to be 80% confident and accurate within 8% of the mean. From a previous survey he found the standard deviation was 12%, how many people would need to be surveyed for this?