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Chapter 10: Problem 61

Construct a \(95 \%\) confidence interval for \(p_{1}-p_{2}\) for the following. $$ n_{1}=100 \quad \hat{p}_{1}=.81 \quad n_{2}=150 \quad \hat{p}_{2}=.77 $$

Short Answer

Expert verified
The 95% confidence interval for \(p_{1}-p_{2}\) is \((-0.07662 , 0.15662)\)

Step by step solution

01

Calculate Point Estimate of Difference

First, calculate the point estimate of the difference. This is simply the difference between the two sample proportions, \(\hat{p}_{1}\) and \(\hat{p}_{2}\). So, \(Point\:Estimate = \hat{p}_{1} - \hat{p}_{2} = 0.81 - 0.77 = 0.04\)
02

Compute Standard Error

Next, calculate the standard error. The formula for the standard error of the difference in proportions is: \(\sqrt{\frac{\hat{p}_{1}(1-\hat{p}_{1})}{n_{1}} + \frac{\hat{p}_{2}(1-\hat{p}_{2})}{n_{2}}}\). Substituting the given values, we find: \(Standard\:Error = \sqrt{\frac{(0.81)(0.19)}{100} + \frac{(0.77)(0.23)}{150}} = 0.0595\)
03

Find Z-score

A 95% confidence interval corresponds to a Z-score of 1.96. This value is obtained from standard statistical tables and is used in the calculation of the confidence interval.
04

Construct Confidence Interval

Finally, construct the confidence interval using the formula: \((Point\:Estimate - Z * Standard\:Error, Point\:Estimate + Z * Standard\:Error)\). This gives us: \((0.04 - 1.96*0.0595, 0.04 + 1.96*0.0595) = (-0.07662 , 0.15662)\). This is the 95% confidence interval for \(p_{1}-p_{2}\)

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Key Concepts

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

Point Estimate
The point estimate is a simple yet powerful concept in statistics, giving us a single value that serves as our best guess for an unknown population parameter. In this context, the point estimate represents the difference between two sample proportions, denoted as \( \hat{p}_{1} \) and \( \hat{p}_{2} \).
First, identify these proportions from your sample data. In our exercise, \( \hat{p}_{1} = 0.81 \) and \( \hat{p}_{2} = 0.77 \). To find the point estimate of their difference:
  • Subtract the second sample proportion from the first: \( 0.81 - 0.77 \).
  • The result is the point estimate, which in this case is \( 0.04 \).
This value serves as an estimate of the difference in the population proportions.
Standard Error
The standard error helps us understand the variability or uncertainty around our point estimate. It measures how much the sample proportion is expected to fluctuate from the population proportion.
To calculate it in the context of comparing two proportions, use the formula:
  • \( \sqrt{\frac{\hat{p}_{1}(1-\hat{p}_{1})}{n_{1}} + \frac{\hat{p}_{2}(1-\hat{p}_{2})}{n_{2}}} \),
  • where \( n_{1} \) and \( n_{2} \) are your sample sizes.
In our specific exercise:
  • Plugging in the values, \( \sqrt{\frac{(0.81)(0.19)}{100} + \frac{(0.77)(0.23)}{150}} \) results in a standard error of approximately \( 0.0595 \).
This number gives us insight into the reliability of the point estimate.
Sample Proportions
Sample proportions are essentially the fractions of the sample that have a particular attribute or characteristic. They are crucial for estimating the parameters of interest in the larger population.
When working with sample proportions, you must:
  • Know the size of your samples, denoted as \( n_{1} \) and \( n_{2} \).
  • Determine the actual proportion of the sample with the characteristic of interest, written as \( \hat{p}_{1} \) and \( \hat{p}_{2} \).
In our exercise, with \( \hat{p}_{1} = 0.81 \) from a sample of 100, and \( \hat{p}_{2} = 0.77 \) from a sample of 150, the sample proportions play a key role in both point estimate and standard error calculations.
Z-score
The Z-score is a measure that tells us how many standard deviations an element is from the mean. It is particularly useful in constructing confidence intervals.
A 95% confidence interval generally corresponds to a Z-score of 1.96. This Z-score is a constant used in conjunction with the standard error to determine the range within which we expect the true difference in population parameters to lie.
To apply the Z-score in our confidence interval calculations:
  • Multiply the standard error by the Z-score (1.96 in this case).
  • Apply it to the point estimate using the formula: \((Point\:Estimate - Z \times Standard\:Error, Point\:Estimate + Z \times Standard\:Error)\).
Concluding from our exercise, the confidence interval should be \((-0.07662, 0.15662)\). This range suggests the likely difference in the actual population proportions with 95% confidence.

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Most popular questions from this chapter

An economist was interested in studying the impact of the recession on dining out, including drive-thru meals at fast food restaurants. A random sample of forty-eight families of four with discretionary incomes between \(\$ 300\) and \(\$ 400\) per week indicated that they reduced their spending on dining out by an average of \(\$ 31.47\) per week, with a sample standard deviation of \(\$ 10.95\). Another random sample of 42 families of five with discretionary incomes between \(\$ 300\) and \(\$ 400\) per week reduced their spending on dining out by an average \(\$ 35.28\) per week, with a sample standard deviation of \(\$ 12.37\). (Note that the two groups of families are differentiated by the number of family members.) Assume that the distributions of reductions in weekly dining-out spendings for the two groups have the same population standard deviation. a. Construct a \(90 \%\) confidence interval for the difference in the mean weekly reduction in diningout spending levels for the two populations. b. Using the \(5 \%\) significance level, can you conclude that the average weekly spending reduction for all families of four with discretionary incomes between \(\$ 300\) and \(\$ 400\) per week is less than the average weekly spending reduction for all families of five with discretionary incomes between \(\$ 300\) and \(\$ 400\) per week?

Gamma Corporation is considering the installation of governors on cars driven by its sales staff. These devices would limit the car speeds to a preset level, which is expected to improve fuel economy. The company is planning to test several cars for fuel consumption without governors for 1 week. Then governors would be installed in the same cars, and fuel consumption will be monitored for another week. Gamma Corporation wants to estimate the mean difference in fuel consumption with a margin of error of estimate of 2 mpg with a \(90 \%\) confidence level. Assume that the differences in fuel consumption are normally distributed and that previous studies suggest that an estimate of \(s_{d}=3 \mathrm{mpg}\) is reasonable. How many cars should be tested? (Note that the critical value of \(t\) will depend on \(n\), so it will be necessary to use trial and error.)

Two local post offices are interested in knowing the average number of Christmas cards that are mailed out from the towns that they serve. A random sample of 80 households from Town A showed that they mailed an average of \(28.55\) Christmas cards with a standard deviation of \(10.30 .\) The corresponding values of the mean and standard deviation produced by a random sample of 58 households from Town B were \(33.67\) and \(8.97\) Christmas cards. Assume that the distributions of the numbers of Christmas cards mailed by all households from both these towns have the same population standard deviation. a. Construct a \(95 \%\) confidence interval for the difference in the average numbers of Christmas cards mailed by all households in these two towns b. Using the \(10 \%\) significance level, can you conclude that the average number of Christmas cards mailed out by all households in Town A is different from the corresponding average for Town B?

Maria and Ellen both specialize in throwing the javelin. Maria throws the javelin a mean distance of 200 feet with a standard deviation of 10 feet, whereas Ellen throws the javelin a mean distance of 210 feet with a standard deviation of 12 feet. Assume that the distances each of these athletes throws the javelin are normally distributed with these population means and standard deviations. If Maria and Ellen each throw the javelin once, what is the probability that Maria's throw is longer than Ellen's?

According to the information given in Exercise \(10.25\), a sample of 45 customers who drive luxury cars showed that their average distance driven between oil changes was 3187 miles with a sample standard deviation of \(42.40\) miles. Another sample of 40 customers who drive compact lower-price cars resulted in an average distance of 3214 miles with a standard deviation of \(50.70\) miles. Suppose that the standard deviations for the two populations are not equal. a. Construct a \(95 \%\) confidence interval for the difference in the mean distance between oil changes for all luxury cars and all compact lower-price cars. b. Using the \(1 \%\) significance level, can you conclude that the mean distance between oil changes is lower for all luxury cars than for all compact lower-price cars? c. Suppose that the sample standard deviations were \(28.9\) and \(61.4\) miles, respectively. Redo parts a and b. Discuss any changes in the results.
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