/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 74 A bank randomly selected 250 che... [FREE SOLUTION] | 91Ó°ÊÓ

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A bank randomly selected 250 checking account customers and found that 110 of them also had savings accounts at the same bank. Construct a \(95 \%\) confidence interval for the true proportion of checking account customers who also have savings accounts.

Short Answer

Expert verified
The 95% confidence interval for the true proportion of checking account customers who also have savings accounts is given by \(p \pm 1.96 * SE\), where p is the sample proportion and SE is the standard error.

Step by step solution

01

Compute the Sample Proportion

First, compute the sample proportion (p) by dividing the number of individuals with both checking and savings accounts (110) by the total number of individuals sampled (250). That is, \(p = 110/250\).
02

Compute the Standard Error

Next, compute the standard error (SE) using the formula: \(SE = sqrt[ p(1-p) / n]\) where n is the sample size (250).
03

Calculate the Confidence Interval

Finally, calculate the 95% confidence interval using the formula: \(p \pm Z(1-\alpha/2) * SE\) Where Z(1-\alpha/2) for a confidence level of 95% is 1.96.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sample Proportion
When conducting surveys or experiments, researchers often need to estimate a population parameter based on a smaller, randomly selected group, known as a sample. A common parameter of interest is the proportion of the population that has a specific characteristic, termed the 'sample proportion'.

Imagine a school where we want to find out how many students are interested in starting a coding club, but it's impractical to ask everyone. Instead, we survey a random group of 100 students, and 25 say 'yes'. Here, the sample proportion, often represented with the symbol \( p \), would be \( p = 25/100 = 0.25 \) or 25%. This gives us a statistical estimate of the true proportion of interested students in the entire school.

In our exercise, the bank determined the sample proportion of customers with both checking and savings accounts by dividing the number of individuals with both accounts (110) by the total sample size (250), which gives us \( p = 110/250 = 0.44 \). This means that 44% of the sampled customers have both types of accounts, and this proportion is used as an estimate of the true proportion for the entire customer base of the bank.
Standard Error
The 'standard error' (SE) reveals how much we expect our sample's statistic, like the sample proportion, to vary from one random sample to another. It's also a crucial element in gauging the precision of our estimate. Consider SE as an alarm system: the smaller it is, the more confident we can be that our sample proportion is close to the real deal.

A high school basketball coach trying to determine the average height of players on future teams may not have the exact heights of potential players. By taking random samples of players' heights over the years, the coach can use SE to predict how much the average height may fluctuate, helping to prepare for different types of players.

The formula for the standard error of a proportion is \( SE = \sqrt{ p(1-p) / n } \), where \( p \) is the sample proportion and \( n \) is the sample size. In our bank example, we calculated the SE to assess how much the sample proportion of customers with both accounts might vary if we took different samples from the population. It serves as the building block for creating a confidence interval around our sample proportion.
Z-score
A 'Z-score' represents how many standard deviations a data point is from the mean of a distribution. It's a way of standardizing scores across different types of data, enabling us to compare apples with oranges. For example, if a student scores 90 on a math test with a mean of 80 and a standard deviation of 10, the Z-score would be 1 - this score is one standard deviation above the mean.

In the context of confidence intervals, the Z-score allows us to decide how far we'd wander from our sample proportion to capture the true population proportion with a certain level of confidence. It's like casting a fishing net of a specific size; the Z-score tells us how big that net is based on the confidence level we want.

For 95% confidence intervals, the Z-score is commonly 1.96, which means we're casting our net to capture the true proportion within 1.96 standard deviations from the sample proportion. In our exercise, employing a Z-score of 1.96 allowed us to calculate the range in which we are 95% confident the true proportion of the bank's customers with both checking and savings accounts lies.

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Most popular questions from this chapter

a. Find the first percentile of Student's \(t\) -distribution with 24 degrees of freedom. b. Find the 95 th percentile of Student's \(t\) -distribution with 24 degrees of freedom. c. Find the first quartile of Student's \(t\) -distribution with 24 degrees of freedom.

A manufacturer of television sets claims that the maintenance expenditures for its product will average no more than \(110\)dollar during the first year following the expiration of the warranty. A consumer group has asked you to substantiate or discredit the claim. The results of a random sample of 50 owners of such television sets showed that the mean expenditure was \(131.60\)dollar and the standard deviation was \(42.46\)dollar At the 0.01 level of significance, should you conclude that the manufacturer's claim is true or not likely to be true?

Karl Pearson once tossed a coin 24,000 times and recorded 12,012 heads. a. Calculate the point estimate for \(p=P(\) head ) based on Pearson's results. b. Determine the standard error of proportion. c. Determine the \(95 \%\) confidence interval estimate for \(p=P(\text { head })\). d. It must have taken Mr. Pearson many hours to toss a coin 24,000 times. You can simulate 24,000 coin tosses using the computer and calculator commands that follow. (Note: A Bernoulli experiment is like a "single" trial binomial experiment. That is, one toss of a coin is one Bernoulli experiment with \(p=0.5;\) and 24,000 tosses of a coin either is a binomial experiment with \(n=24,000\) or is 24,000 Bernoulli experiments. Code: \(0=\) tail, \(1=\) head. The sum of the 1 s will be the number of heads in the 24,000 tosses.) e. How do your simulated results compare with Pearson's? f. Use the commands (part d) and generate another set of 24,000 coin tosses. Compare these results to those obtained by Pearson. Also, compare the two simulated samples to each other. Explain what you can conclude from these results.

It is claimed that the students at a certain university will score an average of 35 on a given test. Is the claim reasonable if a random sample of test scores from this university yields \(33,42,38,37,30,42 ?\) Complete a hypothesis test using \(\alpha=0.05 .\) Assume test results are normally distributed. a. Solve using the \(p\) -value approach. b. Solve using the classical approach.

According to the August 2009 Reader's Digest article "Where Our Garbage Goes," the average American tosses 4.6 pounds of garbage every day. A small town in Vermont initiated a Going Green campaign and asked residents to work on recycling more and reducing their generation of garbage each day. To estimate the average amount of trash discarded by people in their town, 18 households were randomly selected and all were asked to carefully weigh their trash on the same day. The average amount for the sample was 3.89 pounds, with a standard deviation of 1.322 pounds. Is there sufficient evidence that the Vermont town now has significantly lower average daily garbage amounts than the average American household? Use a 0.05 level of significance and assume weights are normally distributed.

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