In a random sample of 765 adults in the United States, 322 say they could not...

In a random sample of 765 adults in the United States, 322 say they could not cover a $400 unexpected expense without borrowing money or going into debt.

A.)Find the mean and the standard deviation of p-hat for this sample

B.) Given the sample data in problem (1) above, construct a confidence interval at the 95% confidence level for the true proportion of Americans who could not cover a $400 unexpected expense without borrowing money or going into debt.

C.)4) Finding confidence intervals stems from getting a normal distribution of a sample proportion p-hat. If I estimate my population proportion to be no greater than 95%, what is the smallest sample size I can use to ensure I have a large enough sample to make a confidence interval?

Please help

Solutions

Expert Solution

orchestra answered 1 year ago

Next > < Previous

Related Solutions

A poll conducted in 2012 asked a random sample of 1358 adults in the United States...

A poll conducted in 2012 asked a random sample of 1358 adults in the United States how much confidence they had in banks and other financial institutions. A total of 149 adults said that they had a great deal of confidence. An economist claims that less than 13% of U.S. adults have a great deal of confidence in banks. Can you conclude that the economist's claim is true? Use both α=0.10 and α=0.01 levels of significance and the P-value method...

The Pew Research Center took a random sample of 2928 adults in the United States in...

The Pew Research Center took a random sample of 2928 adults in the United States in September 2008. In this sample, 53% of 2928 people believed that reducing the spread of acquired immune sample deficiency disease (AIDS) and other infectious diseases was an important policy goal for the U.S. government. If the significance level α=0.05, should we reject or fail to reject the null hypothesis? How would you interpret your results of hypothesis testing? REJECT HO DO NOT REJECT HO...

In the United States, 75 percent of adults wear glasses or contact lenses. A random sample...

In the United States, 75 percent of adults wear glasses or contact lenses. A random sample of 10 adults in the United States will be selected. Which of the following is closest to the probability that fewer than 8 of the selected adults wear glasses or contact lenses?

Among a simple random sample of 322 American adults who do not have a four-year college...

Among a simple random sample of 322 American adults who do not have a four-year college degree and are not currently enrolled in school, 145 said they decided not to go to college because they could not afford school. 1. Calculate a 99% confidence interval for the proportion of Americans who decide to not go to college because they cannot afford it, and interpret the interval in context. Round to 4 decimal places. ( , ) 2. Suppose we wanted...

A sample of n = 1,033 adults in the United States was asked, "Do you think...

A sample of n = 1,033 adults in the United States was asked, "Do you think the police should or should not be allowed to collect DNA information from suspected criminals, similar to how they take fingerprints?" Of those sampled, 68% answered "should". (Round your answers to the nearest tenth.) (a) Calculate the conservative margin of error for the survey, as a percentage. (b) Compute an approximate 95% confidence interval for the percentage of all American adults who think police...

A simple random sample of 101 cities and counties in the United States was selected and...

A simple random sample of 101 cities and counties in the United States was selected and the number of positive COVID-19 cases per 1000 citizens was recorded for each. The mean number of positive cases for this sample of 101 counties and cities was 253, with a standard deviation of 11.4. Due to a couple of cities with extremely high numbers of cases the distribution is skewed heavily to the right. If appropriate, use this information to calculate and interpret...

A simple random sample of birth weights in the United States has a mean of...

A simple random sample of birth weights in the United States has a mean of 3433g. The standard deviation of all birth weights (for the population) is 495g. If this data came from a sample size of 75, construct a 95% confidence interval estimate of the mean birth weight in the United States. If this data came from a sample size of 7,500, construct a 95% confidence interval estimate of the mean birth weight in the United States. Which...

A random sample of 25 workers showed that the average workweek in the United States is...

A random sample of 25 workers showed that the average workweek in the United States is down to only 35 hours, largely because of a rise in part-time workers. The variance of the sample was 18.49 hours. Assuming that the population is normally distributed, calculate a 98% confidence interval for the population standard deviation.

A simple random sample of 35 colleges and universities in the United States has a mean...

A simple random sample of 35 colleges and universities in the United States has a mean tuition of 18700 with a standard deviation of 10800. Construct a 99% confidence interval for the mean tuition for all colleges and universities in the United States. Round the answers to the nearest whole number.

A simple random sample of 40 colleges and universities in the United States has a mean...

A simple random sample of 40 colleges and universities in the United States has a mean tuition of 17800 with a standard deviation of 11000. Construct a 95% confidence interval for the mean tuition for all colleges and universities in the United States. Round the answers to the nearest whole number.

Subjects