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91Ó°ÊÓ

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

An insurance company states that \(90 \%\) of its claims are settled within 30 days. A consumer group selected a random sample of 75 of the company's claims to test this statement. If the consumer group found that 55 of the claims were settled within 30 days, does it have sufficient reason to support the contention that less than \(90 \%\) of the claims are settled within 30 days? Use \(\alpha=0.05\). a. Solve using the \(p\) -value approach. b. Solve using the classical approach.

All tomatoes that a certain supermarket buys from growers must meet the store's specifications of a mean diameter of \(6.0 \mathrm{cm}\) and a standard deviation of no more than \(0.2 \mathrm{cm} .\) The supermarket's buyer visits a potential new supplier and selects a random sample of 36 tomatoes from the grower's greenhouse. The diameter of each tomato is measured, and the mean is found to be 5.94 and the standard deviation is \(0.24 .\) Do the tomatoes meet the supermarket's specs? a. Determine whether an assumption of normality is reasonable. Explain. b. Is the sample evidence sufficient to conclude that the tomatoes do not meet the specs with regard to the mean diameter? Use \(\alpha=0.05\). c. Is the sample evidence sufficient to conclude that the tomatoes do not meet the specs with regard to the standard deviation? Use \(\alpha=0.05\). d. Write a short report for the buyer outlining the findings and recommendations as to whether or not to use this tomato grower to supply tomatoes for sale in the supermarket.

You are interested in comparing the null hypothesis \(p=0.8\) against the alternative hypothesis \(p<0.8 .\) In 100 trials you observe 73 successes. Calculate the \(p\) -value associated with this result.

The May \(30,2008,\) online article "Live with Your Parents After Graduation?" quoted a 2007 survey conducted by Monster-TRAK.com. The survey found that \(48 \%\) of college students planned to live at home after graduation. How large of a sample size would you need to estimate the true proportion of students that plan to live at home after graduation to within \(2 \%\) with \(98 \%\) confidence?

In the past the standard deviation of weights of certain 32.0 -oz packages filled by a machine was 0.25 oz. A random sample of 20 packages showed a standard deviation of 0.35 oz. Is the apparent increase in variability significant at the 0.10 level of significance? Assume package weight is normally distributed. a. Solve using the \(p\) -value approach. b. Solve using the classical approach.

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