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The management of a supermarket chain wanted to investigate if the percentages of men and women who prefer to buy national brand products over the store brand products are different. A sample of 600 men shoppers at the company's supermarkets showed that 246 of them prefer to buy national brand products over the store brand products. Another sample of 700 women shoppers at the company's supermarkets showed that 266 of them prefer to buy national brand products over the store brand products. a. What is the point estimate of the difference between the two population proportions? b. Construct a \(95 \%\) confidence interval for the difference between the proportions of all men and all women shoppers at these supermarkets who prefer to buy national brand products over the store brand products. c. Testing at the \(5 \%\) significance level, can you conclude that the proportions of all men and all women shoppers at these supermarkets who prefer to buy national brand products over the store brand products are different?

Short Answer

Expert verified
The point estimate of the difference between the two population proportions is 0.03. The calculation for the \(95 \%\) confidence interval and the hypothesis test would depend on the calculated standard error and the test statistic respectively. Depending on the results, we will be able to conclude if the proportions of men and women preferring national brand products are significantly different.

Step by step solution

01

Calculation of Point Estimate

The point estimate of the difference between two population proportions is the difference between two sample proportions. For men, the proportion preferring national brand products is \(\frac{246}{600} = 0.41\). For women, it is \(\frac{266}{700} = 0.38\). Hence, point estimate is \(0.41 - 0.38 = 0.03\).
02

Construction of Confidence Interval

Let's denote the proportions and sample sizes as \(p1\), \(p2\), \(n1\), and \(n2\) respectively for better understanding. Here, \(p1 = 0.41\), \(n1 = 600\), \(p2 = 0.38\), and \(n2 = 700\). We will construct a \(95 \%\) confidence interval for the difference between the proportions.First, calculate the standard error \(SE\) using the formula:\[ SE = \sqrt{ \frac{p1(1-p1)}{n1} + \frac{p2(1-p2)}{n2} } \]Then calculate \(E\), the Margin of Error for a \(95\%\) confidence level is \(1.96 \times SE\). The Confidence interval is \((Point \ Estimate - E, Point \ Estimate + E)\).
03

Hypothesis Testing

Finally, we perform a hypothesis testing. For this, the null hypothesis \(H0\) is that the proportions are equal, i.e., \(p1=p2\), and the alternative hypothesis \(Ha\) is \(p1\neq p2\).The test statistic Z is calculated using the formula:\[ Z = \frac{(p1-p2)}{SE} \]Under the null hypothesis, Z follows a standard normal distribution. If the value of Z falls in the rejection region (i.e outside the range formed by -1.96 and +1.96 in this case), we reject the null hypothesis and conclude that the proportions of all men and all women shoppers who prefer to buy national brand products over the store brand products are different.

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Key Concepts

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

Confidence Interval
When conducting hypothesis tests, a confidence interval helps us determine the range within which we expect the true difference in population proportions to lie. It gives us an interval that is constructed around our point estimate. In this scenario, we want to see if the proportion difference of men and women who prefer national brands is significant.
To achieve this, we first need to calculate the standard error (SE) for our estimates. The smaller the SE, the more precise our estimates are. The margin of error (E) is then determined by multiplying the SE by the z-score corresponding to the desired confidence level, which is 1.96 for a 95% confidence interval. Adding and subtracting this margin of error from our point estimate results in the confidence interval.
This interval helps us understand the range of values we might expect if we repeatedly took samples from the same population.
Population Proportions
Population proportions refer to the ratio of individuals in our population with a particular attribute of interest. In this exercise, the focus is on understanding the proportion of men and women who prefer national over store brands.
For men, the observed proportion was calculated as 0.41, while for women, it was 0.38. By comparing these proportions, we can assess whether there's a significant difference in preferences.
The analysis involves using these sample proportions as estimates for their respective population proportions. Given that we work with samples rather than entire populations, it is crucial to be cautious about the reliability of these estimates, which leads us to further validation using standard error and hypothesis testing.
Standard Error
The standard error is a measure of the variability of our sample estimate, which reflects how much the sample proportions might fluctuate from the true population proportions. Calculating it allows us to understand the precision of our estimates of the population difference.
In this exercise scenario, the standard error is calculated using the formula:
  • \[SE = \sqrt{ \frac{p1(1-p1)}{n1} + \frac{p2(1-p2)}{n2} }\]
Here, \(p1\) and \(p2\) are the sample proportions, and \(n1\) and \(n2\) are the sample sizes. This calculation provides us with a foundation to determine the confidence interval and conduct hypothesis testing. A smaller SE indicates that our estimate is more precise, which enhances the reliability of the inference we deduce from our data.
Significance Level
The significance level, often denoted as \(\alpha\), is a threshold we set to decide whether a statistical test result is significant. In this context, we use a 5% (0.05) significance level to test if the proportions of men and women preferring national brands differ.
This level defines our tolerance for incorrectly concluding that there is a difference when there isn't one, also known as a Type I error. By choosing a 5% significance level, we limit the likelihood of such errors to 5%.
Using this threshold, we calculate the test statistic Z from our sample proportions and standard error. We compare this Z value to determine if it falls in the "rejection region," typically bounded by critical values earmarked by the selected significance level. If it lies outside this range, we reject the null hypothesis and conclude that a significant difference exists between the proportions.

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Most popular questions from this chapter

Conduct the following tests of hypotheses, assuming that the populations of paired differences are normally distributed a. \(H_{0}: \mu_{d}=0, \quad H_{1}: \mu_{d} \neq 0, \quad n=26, \quad \bar{d}=9.6, \quad s_{d}=3.9, \quad \alpha=.05\) b. \(H_{0}: \mu_{d}=0, \quad H_{1}: \mu_{d}>0, \quad n=15, \quad \bar{d}=8.8, \quad s_{d}=4.7, \quad \alpha=.01\) c. \(H_{0}: \mu_{d}=0, \quad H_{1}: \mu_{d}<0, \quad n=20, \quad \bar{d}=-7.4, \quad s_{d}=2.3, \quad \alpha=.10\)

A sample of 500 observations taken from the first population gave \(x_{1}=305\). Another sample of 600 observations taken from the second population gave \(x_{2}=348\). a. Find the point estimate of \(p_{1}-p_{2}\). b. Make a \(97 \%\) confidence interval for \(p_{1}-p_{2}\). c. Show the rejection and nonrejection regions on the sampling distribution of \(\hat{p}_{1}-\hat{p}_{2}\) for \(H_{0}: p_{1}=p_{2}\) versus \(H_{1}: p_{1}>p_{2} .\) Use a significance level of \(2.5 \% .\) d. Find the value of the test statistic \(z\) for the test of part \(\mathrm{c}\). e. Will you reject the null hypothesis mentioned in part \(\mathrm{c}\) at a significance level of \(2.5 \%\) ?

Sixty-five percent of all male voters and \(40 \%\) of all female voters favor a particular candidate. A sample of 100 male voters and another sample of 100 female voters will be polled. What is the probability that at least 10 more male voters than female voters will favor this candidate?

The lottery commissioner's office in a state wanted to find if the percentages of men and women who play the lottery often are different. A sample of 500 men taken by the commissioner's office showed that 160 of them play the lottery often. Another sample of 300 women showed that 66 of them play the lottery often. a. What is the point estimate of the difference between the two population proportions? b. Construct a \(99 \%\) confidence interval for the difference between the proportions of all men and all women who play the lottery often. c. Testing at the \(1 \%\) significance level, can you conclude that the proportions of all men and all women who play the lottery often are different?

According to Exercise \(10.27\), an insurance company wants to know if the average speed at which men drive cars is higher than that of women drivers. The company took a random sample of 27 cars driven by men on a highway and found the mean speed to be 72 miles per hour with a standard deviation of \(2.2\) miles per hour. Another sample of 18 cars driven by women on the same highway gave a mean speed of 68 miles per hour with a standard deviation of \(2.5\) miles per hour. Assume that the speeds at which all men and all women drive cars on this highway are both normally distributed with unequal population standard deviations. a. Construct a \(98 \%\) confidence interval for the difference between the mean speeds of cars driven by all men and all women on this highway. b. Test at the \(1 \%\) significance level whether the mean speed of cars driven by all men drivers on this highway is higher than that of cars driven by all women drivers. c. Suppose that the sample standard deviations were \(1.9\) and \(3.4\) miles per hour, respectively. Redo parts a and b. Discuss any changes in the results.

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