Question

In: Statistics and Probability

(2) A survey of households in a small town showed that in 500 of 2,000 sampled...

(2) A survey of households in a small town showed that in 500 of 2,000 sampled households, at least one member attended a town meeting during the year. What is the 95% confidence interval for the proportion of households represented at a town meeting?

Expert Solution

Solution :

Given that,

n = 2000

x = 500

= x / n = 500 /2000 = 0.25

1 - = 1 - 0.25= 0.75

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

Margin of error = E = Z_{/ 2} * $\sqrt$ (( * (1 - )) / n)

= 1.96 * ( $\sqrt$ ((0.25* 0.75) / 2000) = 0.0190

A 95 % confidence interval for population proportion p is ,

- E < P < + E

0.25 - 0.0190 < p < 0.25+ 0.0190

0.2310 < p < 0.2690

The 90% confidence interval for the population proportion p is : ( 0.2310 , 0.2690 )

orchestra answered 2 years ago

In a city of 80,000 households, 800 were randomly sampled in a marketing survey by an...

In a city of 80,000 households, 800 were randomly sampled in a marketing survey by an auto manufacturer. To see how typical this city was, some statistics were collected: the average number of cars per household was 1.32, while the standard deviation for the number of cars per household was 0.8. Construct a 95% confidence interval for the average number of cars per household in the whole city.

A national survey of heads of households showed the percentage of those who asked for a...

A national survey of heads of households showed the percentage of those who asked for a raise and the percentage who got one. According to the survey, of the women interviewed, 26% had asked for a raise, and of those women who had asked for a raise, 45% received the raise. Suppose a woman is randomly selected, then: (a) Find P (woman did not ask for a raise). (b) Find P (woman asked for raise and received raise). You are...

Does it pay to ask for a raise? A national survey of heads of households showed...

Does it pay to ask for a raise? A national survey of heads of households showed the percentage of those who asked for a raise and the percentage who got one. According to the survey, of the women interviewed, 26% had asked for a raise, and of those women who had asked for a raise, 45% received the raise. If a woman is selected at random from the survey population of women, find the following probabilities. (Enter your answers to...

The number of computers per household in a small town Computers 0 1 2 3 Households...

The number of computers per household in a small town Computers 0 1 2 3 Households 298 284 97 18 Find probability distribution Graph using a histogram Describe distribution shape

A recent survey of 200 households in the United States showed that 8 had a single...

A recent survey of 200 households in the United States showed that 8 had a single male as the head of household. A survey of 150 households in Canada showed that 5 had a single male as the head of household. At α =.05, can it be concluded that the proportion in the two countries is different?

A) In 1997, a survey of 820 households showed that 137 of them use e-mail. Use...

A) In 1997, a survey of 820 households showed that 137 of them use e-mail. Use those sample results to test the claim that more than 15% of households use e-mail. Use a 0.05 significance level. Use this information to answer the following questions. What is the P-value? B) In a study of 420,149 cell phone users, 115 subjects developed cancer of the brain or nervous system. Test the claim of a somewhat common belief that such cancers are affected...

In a recent survey of 4276 randomly selected households showed that 3549 of them had telephones....

In a recent survey of 4276 randomly selected households showed that 3549 of them had telephones. Using these results, construct a 99% confidence interval estimate of the true proportion of households with telephones. Use the PANIC acronym

In? 1997, a survey of 840 households showed that 147 of them use? e-mail. Use those...

In? 1997, a survey of 840 households showed that 147 of them use? e-mail. Use those sample results to test the claim that more than? 15% of households use? e-mail. Use a 0.05 significance level. Use this information to answer the following questions. a. Which of the following is the hypothesis test to be? conducted? b. What is the test? statistic? c. What is the? P-value? d. What is the? conclusion? There is not sufficient evidence to support the claim...

In 1997, a survey of 860 households showed that 161 of them use e-mail. Use those...

In 1997, a survey of 860 households showed that 161 of them use e-mail. Use those sample results to test the claim that more than 15% of households use e-mail. Use a 0.05 significance level. What is the P-Value?

A survey of 250 households showed 96 owned at least one snow blower. Find a point...

A survey of 250 households showed 96 owned at least one snow blower. Find a point estimate for p, the population proportion of households that own at least one snow blower.