Question

In: Statistics and Probability

A marketing company survey found that, in the 100 households surveyed, in 65 households the women...

A marketing company survey found that, in the 100 households surveyed, in 65 households the women made the majority of the purchasing decisions. Find a 95% confidence interval for the proportion of households where women made the majority of purchasing decisions. Round your answer to the nearest hundredth.

Expert Solution

Solution :

Given that,

Point estimate = sample proportion = = x / n = 65/100=0.65

1 - = 1- 0.65 =0.35

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 ( Using z table )

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

= 1.96 ( $\sqrt$ ((0.65*0.35) /100 )

E = 0.093

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.65-0.093 < p < 0.65+0.093

(0.557 , 0.743)

orchestra answered 2 years ago

In the survey sponsored by the Lindt chocolate company, 1708 women were surveyed and 85% of...

In the survey sponsored by the Lindt chocolate company, 1708 women were surveyed and 85% of them said that chocolate made them happier. a) Is there anything potentially wrong with this survey? b) Of the 1708 women surveyed, what is the number of them who said that chocolate made them happier? c)Use Excel to construct a 98% confidence interval estimate of the percentage of women who say that chocolate makes them happier. Insert a screenshot, write down the confidence interval...

A company randomly chose 200 US households and found that in 120 of them, the women...

A company randomly chose 200 US households and found that in 120 of them, the women made the majority of the purchasing decisions. We want to know the population proportion of US housholds where women make the majority of purchasing decisions. Construct a 96% confidence interval to find p. a) what is p hat? b) what is alpha? c) what is the value of the test statistic d) what is the value of the standard errror e) what is the...

4. In a survey sponsored by the Lindt chocolate company, 1708 women were surveyed and 85%...

4. In a survey sponsored by the Lindt chocolate company, 1708 women were surveyed and 85% of them said that chocolate made them happier. (a) Is there anything potentially wrong with this survey? (b) Of the 1708 women surveyed, what is the number of them who said that chocolate made them happier? (c) Use Excel to construct a 98% confidence interval estimate of the percentage of women who say that chocolate makes them happier. Insert a screenshot, write down the...

In a survery of 134 households, a Food Marketing Institute found that 68 households spend more...

In a survery of 134 households, a Food Marketing Institute found that 68 households spend more than $125 a week on groceries. Please find the 95% confidence interval for the true proportion of the households that spend more than $125 a week on groceries. A. Enter your answer as an open-interval (i.e., parentheses) using decimals (not percents) accurate to three decimal places. Confidence interval = B. Express the same answer as a tri-linear inequality using decimals (not percents) accurate to...

In a survery of 352 households, a Food Marketing Institute found that 148 households spend more...

In a survery of 352 households, a Food Marketing Institute found that 148 households spend more than $125 a week on groceries. Please find the 98% confidence interval for the true proportion of the households that spend more than $125 a week on groceries. Enter your answer as an open-interval (i.e., parentheses) using decimals (not percents) accurate to three decimal places. Confidence interval = Express the same answer as a tri-linear inequality using decimals (not percents) accurate to three decimal...

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.

1. A company conducted a marketing survey of college students and found that 225 own a...

1. A company conducted a marketing survey of college students and found that 225 own a bicycle and 77 owned a car. If 23 of those surveyed own both a car and a bicycle, how many interviewed have a car or a bicycle? 2. A standard deck of cards consists of four suits (clubs, diamonds, hearts, and spades), with each suit containing 13 cards (ace, two through ten, jack, queen, and king) for a total of 52 cards in all....

A survey of 100 men and 100 women revealed that 15 of the men and 36...

A survey of 100 men and 100 women revealed that 15 of the men and 36 of the women were more than 20% overweight. Prepare a contingency table and test the hypothesis that the two gender groups are homogeneous with respect to being overweight. Please assume an α=.05

In a recent survey of drinking laws, a random sample of 1000 women showed that 65%...

In a recent survey of drinking laws, a random sample of 1000 women showed that 65% were in favor of increasing the legal drinking age. In a random sample of 1000 men , 60% favored increasing the legal drinking age. Test the hypothesis that the percentage of women favoring a higher legal drinking age is greater than the percentage of men. Use a= 0.05 Call women population 1 and men population 2 A) What initial conclusion would be reached in...

Three randomly selected households are surveyed. The numbers of people in the households are 1,3, and...

Three randomly selected households are surveyed. The numbers of people in the households are 1,3, and 8. Assume that samples of size n=2 are randomly selected with replacement from the population of 1,3, and 8. Construct a probability distribution table that describes the sampling distribution of the proportion of odd numbers when samples of sizes n=2 are randomly selected. Does the mean of the sample proportions equal the proportion of odd numbers in the population? Do the sample proportions target...