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Chapter 10: Problem 71

A state that requires periodic emission tests of cars operates two emissions test stations, \(\mathrm{A}\) and \(\mathrm{B}\), in one of its towns. Car owners have complained of lack of uniformity of procedures at the two stations, resulting in different failure rates. A sample of 400 cars at Station A showed that 53 of those failed the test; a sample of 470 cars at Station B found that 51 of those failed the test. 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 two population proportions. c. Testing at the \(5 \%\) significance level, can you conclude that the two population proportions are different? Use both the critical-value and the \(p\) -value approaches.

Short Answer

Expert verified
a. The point estimate of the difference between the two population proportions can be determined by subtracting the failure rate at Station B from the failure rate at Station A. b. A 95% confidence interval for the difference between the two population proportions can be established using the point estimate 卤 1.96 times the standard error. c. One can determine if the two population proportions differ significantly by making a comparison of the calculated z-value and p-value to their respective thresholds at a 5% significance level.

Step by step solution

01

Calculate Failure Rates

Start by calculating the failure rates at each station. For Station A, divide the number of failed tests (53) by the total number of tests (400). Call this proportion \(p_{A}\). For Station B, divide the number of failed tests (51) by the total number (470). Call this proportion \(p_{B}\).
02

Point Estimate of Difference

Next, determine the point estimate of the difference between the two population proportions. Subtract \(p_{B}\) from \(p_{A}\). This difference, referred denoted as \(p_{D}\), represents the point estimate of the difference.
03

Constructing 95% Confidence Interval

By calculating the standard error and using the Z-value for a 95% confidence interval (which is 1.96), construct a 95% confidence interval for the difference between the two population proportions. The interval is given by \(p_{D} \pm 1.96 * SE\) where \(SE\) is the standard error calculated as \(\sqrt{ p_{A}(1 - p_{A})/n_{A} + p_{B}(1 - p_{B})/n_{B} }\).
04

Hypothesis Test

Lastly, perform a hypothesis test to see if the two proportions are significantly different. Begin by stating the null hypothesis that the proportions are equal, i.e., \(p_{A} = p_{B}\). Then, compute the z-value and compare it to the critical z-value for a 5% significance level (which is 1.96). Also calculate the p-value (probability of obtaining a result as extreme as the observed one under the null hypothesis) and compare it to the significance level (0.05).

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Key Concepts

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

Difference of Proportions
In statistical analysis, understanding the difference between proportions involves comparing two distinct groups to find out if there's a significant difference between them. For example, in this case, we are evaluating the emission test failure rates at two different stations. This technique is crucial when you have binary outcomes (e.g., pass or fail) from two independent samples and want to see if the difference in their proportions is notable.

When calculating the difference of proportions, follow these steps:
  • Calculate the proportion for each group. This is done by dividing the number of "successes" (or in this case, failures) by the total number of observations in each group.
  • The difference of proportions is then found by subtracting one group's proportion from the other.
This difference provides a point estimate of how much the proportions deviate from each other, which can then be further analyzed through confidence intervals and hypothesis testing.
Confidence Interval
A confidence interval offers a range within which you expect the true difference of proportions to lie, based on your sample data. It is an essential concept because it incorporates a margin of error, giving you a way to gauge the reliability of your estimate.

To construct a confidence interval for the difference of proportions:
  • Determine the point estimate of the difference of proportions, which is the previously calculated difference.
  • Calculate the standard error, which helps to quantify the accuracy of your estimate.
  • Apply a Z-value based on your desired confidence level (e.g., 1.96 for 95%).
The 95% confidence interval, for example, would be calculated as the point estimate 卤 (1.96 * standard error). This means you are 95% confident that the true difference between the populations' proportions lies within this interval.
Significance Level
The significance level, often denoted by \(\alpha\), is a threshold used in hypothesis testing to decide whether to reject the null hypothesis. It helps you determine the strength of your evidence. In this problem, the significance level is set at 5%, or \(\alpha = 0.05\).

A 5% significance level implies:
  • There is a 5% risk of concluding that there's a difference when there isn't one (Type I error).
  • This threshold sets what is considered a "rare" event under the null hypothesis. Anything more extreme leads to the rejection of the null hypothesis.
  • It is crucial to compare the p-value from your test to this significance level to decide on the hypothesis.
By maintaining a standard significance level, you ensure consistency across statistical analyses, which helps validate the reliability of your study's conclusions.
Standard Error
The standard error measures the statistical accuracy of an estimate. It represents the spread or variability of a sample proportion estimate. In hypothesis testing for proportions, the standard error helps assess how much the estimate might vary due to sampling error.

To compute the standard error for the difference of proportions between two samples:
  • Use the formula \(\sqrt{p_{A}(1 - p_{A})/n_{A} + p_{B}(1 - p_{B})/n_{B}}\), where \(p_{A}\) and \(p_{B}\) are the sample proportions, and \(n_{A}\) and \(n_{B}\) are the sample sizes for stations A and B, respectively.
  • Smaller standard error values suggest that the difference in proportions is likely more precise and reliable.
Understanding standard error allows you to gauge the margin of error and the stability of your confidence intervals or hypothesis test results.

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Most popular questions from this chapter

A company that has many department stores in the southern states wanted to find at two such stores the percentage of sales for which at least one of the items was returned. A sample of 800 sales randomly selected from Store A showed that for 280 of them at least one item was returned. Another sample of 900 sales randomly selected from Store B showed that for 279 of them at least one item was returned. a. Construct a \(98 \%\) confidence interval for the difference between the proportions of all sales at the two stores for which at least one item is returned. b. Using the \(1 \%\) significance level, can you conclude that the proportions of all sales for which at least one item is returned is higher for Store A than for Store \(B\) ?

The following information was obtained from two independent samples selected from two populations with unknown but equal standard deviations. $$ \begin{array}{lll} n_{1}=55 & \bar{x}_{1}=90.40 & s_{1}=11.60 \\ n_{2}=50 & \bar{x}_{2}=86.30 & s_{2}=10.25 \end{array} $$ a. What is the point estimate of \(\mu_{1}-\mu_{2}\) ? b. Construct a \(99 \%\) confidence interval for \(\mu_{1}-\mu_{2}\).

In a random sample of 800 men aged 25 to 35 years, \(24 \%\) said they live with one or both parents. In another sample of 850 women of the same age group, \(18 \%\) said that they live with one or both parents. a. Construct a \(95 \%\) confidence interval for the difference between the proportions of all men and all women aged 25 to 35 years who live with one or both parents. b. Test at the \(2 \%\) significance level whether the two population proportions are different. c. Repeat the test of part b using the \(p\) -value approach.

Maine Mountain Dairy claims that its 8-ounce low-fat yogurt cups contain, on average, fewer calories than the 8-ounce low-fat yogurt cups produced by a competitor. A consumer agency wanted to check this claim. A sample of 27 such yogurt cups produced by this company showed that they contained an average of 141 calories per cup. A sample of 25 such yogurt cups produced by its competitor showed that they contained an average of 144 calories per cup. Assume that the two populations are normally distributed with population standard deviations of \(5.5\) and \(6.4\) calories, repectively. a. Make a \(98 \%\) confidence interval for the difference between the mean number of calories in the 8-ounce low-fat yogurt cups produced by the two companies. b. Test at the \(1 \%\) significance level whether Maine Mountain Dairy's claim is true. c. Calculate the \(p\) -value for the test of part b. Based on this \(p\) -value, would you reject the null hypothesis if \(\alpha=.005 ?\) What if \(\alpha=.025\) ?

Manufacturers of two competing automobile models, Gofer and Diplomat, each claim to have the lowest mean fuel consumption. Let \(\mu_{1}\) be the mean fuel consumption in miles per gallon (mpg) for the Gofer and \(\mu_{2}\) the mean fuel consumption in mpg for the Diplomat. The two manufacturers have agreed to a test in which several cars of each model will be driven on a 100 -mile test run. Then the fuel consumption, in mpg, will be calculated for each test run. The average of the mpg for all 100 -mile test runs for each model gives the corresponding mean. Assume that for each model the gas mileages for the test runs are normally distributed with \(\sigma=2 \mathrm{mpg}\). Note that each car is driven for one and only one 100 -mile test run. a. How many cars (i.e., sample size) for each model are required to estimate \(\mu_{1}-\mu_{2}\) with a \(90 \%\) confidence level and with a margin of error of estimate of \(1.5 \mathrm{mpg}\) ? Use the same number of cars (i.e., sample size) for each model. b. If \(\mu_{1}\) is actually \(33 \mathrm{mpg}\) and \(\mu_{2}\) is actually \(30 \mathrm{mpg}\), what is the probability that five cars for each model would yield \(\bar{x}_{1} \geq \bar{x}_{2}\) ?
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