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Chapter 10: Problem 71

A state that requires periodic emission tests of cars operates two emission 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 \(\mathrm{B}\) 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 a \(5 \%\) 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 is approximately -0.003.\\ b. The 95\% confidence interval for the difference in proportions is approximately (-0.02, 0.01).\\ c. Depending on the calculated p-value if it鈥檚 greater than 0.05 and z-score if it鈥檚 within the critical region, the decision will be made.

Step by step solution

01

Calculate Point Estimate

To calculate the point estimate of the difference between the two population proportions, subtract the proportion of failures in Station B from the proportion of failures in Station A. The proportions are calculated as the number of failures divided by the sample size for each station. The formula to use is \(p_A - p_B\) where \(p_A = 53/400\) for Station A and \(p_B = 51/470\) for Station B.
02

Construct a 95\% Confidence Interval

To find a 95\% confidence interval for the difference between the two population proportions, use the following formula:\[(p_A - p_B) \pm Z *\sqrt{\frac{p_A(1-p_A)}{n_A} + \frac{p_B(1-p_B)}{n_B}}\]Where \(Z\) is the z-score corresponding to the desired confidence level (for a 95\% confidence level, \(Z\) is approximately 1.96), \(n_A\) and \(n_B\) are the sample sizes for Station A (400) and Station B (470) respectively.
03

Hypothesis Testing

To test whether there is a significant difference between the two proportions at a 5\% significance level, perform a two-sample z-test for proportions. Set up the null and alternative hypotheses:Null hypothesis, \(H_0: p_A = p_B\) Alternative hypothesis, \(H_1: p_A \neq p_B\)Calculate the z-score using below formula,\[z = \frac{(p_A - p_B)}{\sqrt{p(1-p)(\frac{1}{n_A} + \frac{1}{n_B})}\]where \(p\) is the pooled sample proportion, calculated as \((success\_A+success\_B) / (n_A+n_B)\)And calculate the p-value which will be compared to the significance level. If the p-value is less than 0.05, reject the null hypothesis. Compare the calculated z value to the critical z value (for a 5\% significance level, the critical z value is approximately 卤1.96). If the calculated z falls in the critical region, Reject null hypothesis.

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Key Concepts

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

Point Estimate
To tackle any hypothesis testing task like the one described, it's important to understand the idea of a point estimate. A point estimate gives us a single value that is the best guess of a population parameter based on sample data. In this context, we're estimating the difference between two population proportions.

For example, at emission test Stations A and B, we see failure rates. For Station A, 53 out of 400 cars failed, so the proportion of failures, or point estimate, is \( p_A = \frac{53}{400} \). Similarly, for Station B, we have \( p_B = \frac{51}{470} \).
  • The point estimate of the difference of proportions is simply the subtraction: \( p_A - p_B \).
  • This estimate gives us an intuitive sense of how the two stations compare in terms of their test failure rates.
Calculating point estimates is a crucial starting point in hypothesis testing, as it provides a foundation for further statistical analysis.
Confidence Interval
After having the point estimate, it's beneficial to construct a confidence interval around that estimate. A confidence interval gives a range of values that is believed to contain the population parameter with a certain level of confidence. Here, we construct a 95% confidence interval, meaning we can say, with 95% certainty, that the interval includes the true difference between the population proportions.

The confidence interval is calculated using:
  • The point estimate \((p_A - p_B)\).
  • The z-score for 95% confidence which is approximately 1.96.
  • The variability from the samples, captured by a formula that accounts for both sample sizes and proportions.
The formula used is:\[ (p_A - p_B) \pm Z * \sqrt{\frac{p_A(1-p_A)}{n_A} + \frac{p_B(1-p_B)}{n_B}} \]Here, \(Z\) is the z-score that reflects how confident we want to be about our interval, having already calculated this interval helps us understand the reliability of our comparison.
Z-test
Once we have our point estimate and confidence interval, we can further explore whether the differences are statistically significant using a Z-test. A Z-test is a type of hypothesis test used when comparing proportions and other statistical measures sampled from populations that are normally distributed.

In this scenario, the goal is to determine if there is a statistically significant difference between the proportions of failures from Stations A and B, at a 5% significance level:
  • The null hypothesis \( H_0: p_A = p_B \) implies no difference exists between stations.
  • The alternative hypothesis \( H_1: p_A eq p_B \) suggests a noteworthy difference.
We calculate a z-score using the formula:\[ z = \frac{(p_A - p_B)}{\sqrt{p(1-p)(\frac{1}{n_A} + \frac{1}{n_B})}} \]where \( p \) is the pooled sample proportion.
By comparing the resulting z-score to critical z-values or using a p-value, we assess the validity of our null hypothesis.
Significance Level
A significance level is pivotal in hypothesis testing as it represents the probability of rejecting the null hypothesis when it is actually true. In other words, it is the maximum risk we are willing to take of making a wrong decision.

In this situation, the significance level is set at 5% (or 0.05), which is a common threshold in statistical testing. If the calculated p-value from our Z-test is less than the significance level, we have grounds to reject the null hypothesis, suggesting there's evidence that the population proportions are indeed different.
  • The significance level sets the standard for what is considered statistically significant.
  • It helps us decide whether the difference in sample proportions is due to chance or represents a real difference in the populations.
Understanding the significance level is crucial for interpreting the results of your hypothesis test correctly, ensuring informed decisions based on statistical evidence.

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Most popular questions from this chapter

A consumer organization tested two paper shredders, the Piranha and the Crocodile, designed for home use. Each of 10 randomly selected volunteers shredded 100 sheets of paper with the Piranha, and then another sample of 10 randomly selected volunteers each shredded 100 shects with the Crocodile. The Piranha took an average of 203 seconds to shred 100 sheets with a standard deviation of 6 seconds. The Crocodile took an average of 187 seconds to shred 100 sheets with a standard deviation of 5 seconds. Assume that the shredding times for both machines are normally distributed with equal but unknown standard deviations. a. Construct a \(99 \%\) confidence interval for the difference between the two population means. b. Using a \(1 \%\) significance level, can you conclude that the mean time taken by the Piranha to shred 100 sheets is higher than that for the Crocodile? c. What would your decision be in part \(\mathrm{b}\) if the probability of making a Type 1 error were zero? Explain.

Find the following confidence intervals for \(\mu_{d}\), assuming that the populations of paired differences are normally distributed. a. \(n=11, \quad \bar{d}=25.4, \quad s_{d}=13.5, \quad\) confidence level \(=99 \%\) b. \(n=23, \quad \bar{d}=13.2, \quad s_{d}=4.8, \quad\) confidence level \(=95 \%\) c. \(n=18, \quad \bar{d}=34.6, \quad s_{d}=11.7\), confidence level \(=90 \%\)

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 \(m p g\) 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.c., 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 \(\bar{x}_{1} \geq \bar{x}_{2}\) ?

Employees of a large corporation are concerned about the declining quality of medical services provided by their group health insurance. A random sample of 100 office visits by employees of this corporation to primary care physicians during 2004 found that the doctors spent an average of 19 minutes with each patient. This year a random sample of 108 such visits showed that doctors spent an average of \(15.5\) minutes with each patient. Assume that the standard deviations for the two populations are \(2.7\) and \(2.1\) minutes, respectively. a. Construct a \(95 \%\) confidence interval for the difference between the two population means for these two years. b. Using a \(2.5 \%\) level of significance, can you conclude that the mean time spent by doctors with each patient is lower for this year than for \(2004 ?\) c. What would your decision be in part \(\mathrm{b}\) if the probability of making a Type I error were zero? Explain.

The Bath Heritage Days, which take place in Bath, Maine, have been popular for, among other things, an cating contest. In 2009 , the contest switched from blueberry pie to a Whoopie Pie, which consists of two large, chocolate cake-like cookies filled with a large amount of vanilla cream. Suppose the contest involves eating nine Whoopie Pies, each weighing \(1 / 3\) pound. The following data represent the times (in seconds) taken by cach of the 13 contestants (all of whom finished all nine Whoopie Pies) to eat the first Whoopie Pie and the last (ninth) Whoopie Pie. \begin{tabular}{l|rrrrrrrrrrrrr} \hline Contestant & \(\mathbf{1}\) & \(\mathbf{2}\) & \(\mathbf{3}\) & \(\mathbf{4}\) & \(\mathbf{5}\) & \(\mathbf{6}\) & \(\mathbf{7}\) & \(\mathbf{8}\) & \(\mathbf{9}\) & \(\mathbf{1 0}\) & \(\mathbf{1 1}\) & \(\mathbf{1 2}\) & \(\mathbf{1 3}\) \\ \hline First pie & 49 & 59 & 66 & 49 & 63 & 70 & 77 & 59 & 64 & 69 & 60 & 58 & 71 \\ \hline Last pie & 49 & 74 & 92 & 93 & 91 & 73 & 103 & 59 & 85 & 94 & 84 & 87 & 111 \\ \hline \end{tabular} a. Make a 95\% confidence interval for the mean of the population paired differences, where a paired difference is equal to the time taken to eat the ninth pie (which is the last pie) minus the time taken to cat the first pie. b. Using a \(10 \%\) significance level, can you conclude that the average time taken to eat the ninth pie (which is the last pie) is at least 15 seconds more than the average time taken to eat the first pie.
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