Question

1.A survey of households in a small town showed that in 850 of 1,200 sampled households, at least one member attended a town meeting during the year. Using the 99% level of confidence, what is the confidence interval for the proportion of households represented at a town meeting?

2.A survey of an urban university (population of 25,450) showed that 750 of 1,100 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?

3.A survey of 25 grocery stores revealed that the average price of a gallon of milk was $2.98, with a standard error of $0.10. What is the 98% confidence interval to estimate the true cost of a gallon of milk?

4.A student wanted to construct a 95% confidence interval for the mean age of students in her statistics class. She randomly selected nine students. Their mean age was 19.1 years with a sample standard deviation of 1.5 years. What is the 99% confidence interval for the population mean?

Answer #1

1)

Sample size = n = 1200

x = 850

Sample proportion is

We have to construct 99% confidence interval for the population proportion.

Formula is

Here E is a margin of error.

Zc = 2.58

So confidence interval is ( 0.7083 - 0.0339 , 0.7083 + 0.0339)
=> **( 0.6745 , 0.7422)**

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For the next questions please repost!

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