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Marketing: Customer Loyalty In a marketing survey, a random sample of 730 women shoppers revealed that 628 remained loyal to their favorite supermarket during the past year (i.e., did not switch stores) (Source: Trends in the United States: Consumer Attitudes and the Supermarket, The Research Department, Food Marketing Institute). (a) Let \(p\) represent the proportion of all women shoppers who remain loyal to their favorite supermarket. Find a point estimate for \(p\). (b) Find a \(95 \%\) confidence interval for \(p .\) Give a brief explanation of the meaning of the interval. (c) Interpretation As a news writer, how would you report the survey results regarding the percentage of women supermarket shoppers who remained loyal to their favorite supermarket during the past year? What is the margin of error based on a \(95 \%\) confidence interval?

Step by step solution

01

Point Estimate for p

To find a point estimate for the proportion \( p \), we divide the number of women who remained loyal to their supermarket by the total number of women surveyed. In this case, we calculate it as follows: \( \hat{p} = \frac{628}{730} = 0.8603 \). Thus, the point estimate for \( p \) is 0.8603.

02

Calculate Standard Error

The standard error (SE) for the proportion \( \hat{p} \) is calculated using the formula \( SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \), where \( n \) is the sample size. Substituting the values, we get \( SE = \sqrt{\frac{0.8603 \times 0.1397}{730}} = 0.012 \).

03

Find the 95% Confidence Interval

To find the 95% confidence interval, we use the formula \( \hat{p} \pm Z \times SE \), where \( Z \) is the Z-value for a 95% confidence level, approximately 1.96. Thus, the interval is \( 0.8603 \pm 1.96 \times 0.012 \). Calculating this gives the interval \( (0.8363, 0.8843) \).

04

Explain the Confidence Interval

The confidence interval \((0.8363, 0.8843)\) means that we are 95% confident that the true proportion of all women shoppers who remain loyal to their favorite supermarket lies between 83.63% and 88.43%. This considers the uncertainty inherent in estimating a population parameter from a sample.

05

Report the Survey Results

As a news writer, you might report: "According to a recent survey, it is estimated that between 83.63% and 88.43% of women shoppers remained loyal to their favorite supermarket over the past year."

06

Determine the Margin of Error

The margin of error (MOE) at a 95% confidence level is half the width of the confidence interval. Calculating this gives \( MOE = (0.8843 - 0.8363) / 2 = 0.024 \), or 2.4%.

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Key Concepts

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

Point Estimation

Point estimation is a process used in statistics to provide the most accurate value for a population parameter based on sample data. In this case, we are looking to estimate the proportion of all women shoppers who remained loyal to their favorite supermarkets using a sample. The sample consists of 730 women, of which 628 remained loyal. The point estimate for the population proportion, which we denote as \( \hat{p} \), is calculated by dividing the number of loyal shoppers by the total number in the sample.

• The formula for calculating the point estimate \( \hat{p} \) is \( \hat{p} = \frac{628}{730} \).
• This results in \( \hat{p} = 0.8603 \), which implies that roughly 86% of the women in the sample remained loyal.

Thus, in the context of our sample survey, 86.03% of the surveyed women did not switch their preferred supermarket over the past year.

Confidence Interval

Constructing a confidence interval enables statisticians to estimate a range within which the true population parameter (in this case, the proportion \( p \) of all loyal female shoppers) is likely to lie. The confidence interval provides an interval estimate, compared to the single value estimate from point estimation. The 95% confidence interval tells us about the reliability of our point estimate.

• The calculation involves the formula \( \hat{p} \pm Z \times SE \), where \( Z \) is the critical value corresponding to the desired confidence level (1.96 for 95%).
• Using our computed point estimate \( \hat{p} = 0.8603 \) and the standard error \( SE = 0.012 \), the interval calculation becomes \( 0.8603 \pm 1.96 \times 0.012 \).
• This gives us the interval \((0.8363, 0.8843)\).

In simple terms, with 95% confidence, we claim that the true proportion of all women shoppers remaining loyal lies between 83.63% and 88.43%.

Margin of Error

The margin of error is a crucial part of the confidence interval as it indicates the degree of sampling error in a survey's results. It helps to understand how much error might affect the estimates due to using a sample rather than a whole population.

• The margin of error is calculated from half the width of the confidence interval.
• In this survey, it is calculated as \( \frac{0.8843 - 0.8363}{2} = 0.024 \), which translates to a 2.4% margin of error.

This means that the estimated proportion (86.03%) could vary by plus or minus 2.4% due to sample variability when making inferences about the whole population. Therefore, we can say the proportion of loyal women shoppers is 86.03% with a possible error margin of 2.4%.