Question

Suppose that we wish to assess whether more than 60 percent of all U.S. households in...

Suppose that we wish to assess whether more than 60 percent of all U.S. households in a particular income class bought life insurance last year. That is, we wish to assess whether p, the proportion of all U.S. households in the income class that bought life insurance last year, exceeds .60. Assume that an insurance survey is based on 1,000 randomly selected U.S. households in the income class and that 640 of these households bought life insurance last year. a) Assuming that p equals .60 and the sample size is 1,000, what is the probability of observing a sample proportion that is at least .64? b) Based on your answer in part a, do you think more than 60 percent of all U.S. households in the income class bought life insurance last year? Explain.

Homework Answers

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
Suppose that we wish to assess whether more than 60 percent of all U.S. households in...
Suppose that we wish to assess whether more than 60 percent of all U.S. households in a particular income class bought life insurance last year. That is, we wish to assess whether p, the proportion of all U.S. households in the income class that bought life insurance last year, exceeds .60. Assume that an insurance survey is based on 1,000 randomly selected U.S. households in the income class and that 640 of these households bought life insurance last year. 1....
Respond to the following in a minimum of 175 words: Suppose that we wish to assess...
Respond to the following in a minimum of 175 words: Suppose that we wish to assess whether more than 60 percent of all U.S. households in a particular income class bought life insurance last year. That is, we wish to assess whether p, the proportion of all U.S. households in the income class that bought life insurance last year, exceeds .60. Assume that an insurance survey is based on 1,000 randomly selected U.S. households in the income class and that...
Suppose a national news agency reported that more than 88% of all households purchased at least...
Suppose a national news agency reported that more than 88% of all households purchased at least one item online last year. We wish to test this statement and randomly sample 200 households. When asked if they made any online purchases last year 180 reported “YES”. At the 0.05 level of significance, does the sample data support the statement made by the national news agency? 1. State the null and alternative hypotheses of interest. Ho ___________________________ HA ___________________________ 2. What is...
uppose a national news agency reported that more than 88% of all households purchased at least...
uppose a national news agency reported that more than 88% of all households purchased at least one item online last year. We wish to test this statement and randomly sample 200 households. When asked if they made any online purchases last year 180 reported “YES”. At the 0.05 level of significance, does the sample data support the statement made by the national news agency? 1. State the null and alternative hypotheses of interest. Ho ___________________________ HA ___________________________ 2. What is...
Research shows that even for married couples in two-income households, women perform significantly more housework than...
Research shows that even for married couples in two-income households, women perform significantly more housework than men (see Hersh and Stratton 2000). You wish to replicate this study in 2019 and you sample populations of married men and women, all of whom work fulltime outside the home. Housework (Y) is measured in hours and the standard deviations are significantly different from one another. Do your data below support the previous research findings (that women perform more housework) Women Men 1...
Thirty-eight percent of all Americans drink bottled water more than once a week (Natural resources Defense...
Thirty-eight percent of all Americans drink bottled water more than once a week (Natural resources Defense Council, December 4, 2015). Suppose you have been hired by the Natural Resources Defence Council to investigate bottled water consumption in St. Paul. You plan to select a sample of St. Paulites to estimate the proportion who drink bottled water more than once a week. Assume the population proportion of St. Paulites who drink bottled water more than once a week is , the...
1. We wish to estimate what percent of adult residents in a certain county are parents....
1. We wish to estimate what percent of adult residents in a certain county are parents. Out of 400 adult residents sampled, 296 had kids. Based on this, construct a 95% confidence interval for the proportion p of adult residents who are parents in this county. Express your answer in tri-inequality form. Give your answers as decimals, to three places. __< p <__ Express the same answer using the point estimate and margin of error. Give your answers as decimals,...
Absentee rates - Friday vs Wednesday: We want to test whether or not more students are...
Absentee rates - Friday vs Wednesday: We want to test whether or not more students are absent on Friday afternoon classes than on Wednesday afternoon classes. In a random sample of 302 students with Friday afternoon classes, 48 missed the class. In a different random sample of 307 students with Wednesday afternoon classes, 32 missed the class. The table below summarizes this information. The standard error (SE) is given to save calculation time if you are not using software. Data...
Absentee rates - Friday vs Wednesday: We want to test whether or not more students are...
Absentee rates - Friday vs Wednesday: We want to test whether or not more students are absent on Friday afternoon classes than on Wednesday afternoon classes. In a random sample of 302 students with Friday afternoon classes, 48 missed the class. In a different random sample of 307 students with Wednesday afternoon classes, 30 missed the class. The table below summarizes this information. The standard error (SE) is given to save calculation time if you are not using software. Data...
1. A poll of 2234 U.S. adults found that 31% regularly used Facebook as a news...
1. A poll of 2234 U.S. adults found that 31% regularly used Facebook as a news source. Find the margin of error and confidence interval for the percentage of U.S. adults who regularly use Facebook as a news source, at the 90% level of confidence. Round all answers to 2 decimal places. Margin of Error (as a percentage): Confidence Interval: % to % Find the margin of error and confidence interval for the percentage of U.S. adults who regularly use...