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An insurance company wants to know if the average speed at which men drive cars is greater 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 the same population standard deviation. 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 a \(1 \%\) significance level whether the mean speed of cars driven by all men drivers on this highway is greater than that of cars driven by all women drivers.

Short Answer

Expert verified
The 98% confidence interval for the difference of the mean speeds is \(4 ± 2.33 * s_{d}\). To test the hypothesis at a 1% significance level, we calculate a Z-score and compare it with the critical Z value. If the calculated Z exceeds the critical Z, we reject the null hypothesis and conclude that the mean speed of cars driven by men is significantly greater than the mean speed of cars driven by women on this highway.

Step by step solution

01

Identify Information

First, we need to write down all the numbers from the given problem. \n\nFor men drivers, the parameters are: sample size n1 = 27, sample mean \(\mu_{1} = 72\), and sample standard deviation \(s_{1}= 2.2\). \nFor women drivers, the parameters are: sample size n2 = 18, sample mean \(\mu_{2} = 68\), and sample standard deviation \(s_{2}= 2.5\). \n\nThe confidence level for the interval is: \(\alpha = 0.98\) or \(98 \%\). \n\nThe confidence level for the test is: \(\alpha = 0.01\) or \(1 \%\).
02

Formulate Hypotheses

For part b) we should formulate the null hypothesis \(H_{0}\) and the alternative hypothesis \(H_{1}\). \n\n\(H_{0}: \mu_{1} = \mu_{2}\) - The null hypothesis is that the speeds are the same for men and women drivers. \n\n\(H_{1}: \mu_{1} > \mu_{2}\) - The alternative hypothesis is that men drivers have a higher average speed than women drivers.
03

Calculate Average Difference And Standard Deviation Difference

We calculate the average difference: \(\bar{x}_{d} = \mu_{1} - \mu_{2} = 72 - 68 = 4\). \n\nNext, we calculate the standard deviation of the difference: \(s_{d} = \sqrt{ \frac{s_{1}^2}{n_{1}} + \frac{s_{2}^2}{n_{2}} } = \sqrt{ \frac{2.2^2}{27} + \frac{2.5^2}{18}}.\)
04

Find Z Score For 98% Confidence Interval

To find the z-score corresponding to a 98% confidence interval, we need to look it up in a z-table or use a statistical calculator. The z-score is approximately ±2.33 given that we have to divide the remaining 2% area evenly on both sides.
05

Construct Confidence Interval

The confidence interval can be calculated as follows: \(\bar{x}_{d} ± z * s_{d}\). \n\nSubstituting the values, we have: \(4 ± 2.33 * s_{d}\).
06

Conduct Hypothesis Test

To test the hypotheses, we calculate the test statistic (Z) as follows: \n\n\(Z = \frac{ \bar{x}_{d} - 0 }{ s_{d} }\). \n\nWe compare this Z value with the critical Z value from the Z table for a 1% significance level. If our calculated Z exceeds the critical Z, we reject \(H_{0}\) and conclude that the mean speed of cars driven by men is significantly greater than that driven by women on this highway.

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Key Concepts

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

Confidence Interval
A confidence interval provides a range of values that is likely to contain the true difference between the means of two populations. In hypothesis testing, it's crucial to determine how confident we can be about our inferences regarding population parameters based on sample statistics. The confidence interval is determined by the sample mean, the standard deviation of the sample, and a critical value from the standard normal distribution corresponding to the desired confidence level.

For example, if you want to construct a 98% confidence interval for the difference in mean speeds between cars driven by men and those driven by women, follow these steps:
  • Calculate the sample means and standard deviations for both groups.
  • Find the average difference between the means.
  • Use the z-score associated with a 98% confidence level, which is approximately ±2.33, to determine the range.
  • Compute the standard deviation of the difference to adjust the interval width.
By calculating the interval as \[\text{mean difference} \pm z \times \text{standard deviation of} \; \text{difference}\]we obtain a range within which we expect the true difference in mean speeds to lie with 98% certainty.
Sample Mean
The sample mean is a central value obtained by adding all the values of a sample and dividing by the number of observations. It serves as an estimate of the population mean. In our exercise, you have two groups: cars driven by men and cars driven by women, each having its own sample mean.

To calculate the sample mean for men, you take the collective speeds of cars driven by men and divide by the number of cars sampled, which gives a mean speed of 72 mph. Similarly, for women, the calculated sample mean is 68 mph.
The sample mean allows researchers and analysts to make inferences about larger populations from smaller sample data. Understanding the sample mean is vital because it acts as a building block for further statistical analysis, such as constructing confidence intervals or performing hypothesis tests.
Standard Deviation
Standard deviation is a measure of how spread out the numbers in a data set are. It tells you about the variability of speeds among the drivers in each group. A smaller standard deviation means the data points are close to the sample mean, while a larger standard deviation indicates more variability.

In the given problem, the standard deviation is 2.2 mph for men and 2.5 mph for women. These values indicate how much each group's driving speeds tend to deviate from their respective mean speeds.

Standard deviation is crucial for constructing confidence intervals and performing hypothesis tests because it affects the width of the confidence interval and the calculation of the test statistic.
Z-Score
A z-score is a statistical term indicating how many standard deviations a data point or sample statistic is from the mean. In the context of hypothesis testing, a z-score is used to compare the difference between sample means to the variability measured by the standard deviation of that difference.

For instance, when constructing a confidence interval or performing hypothesis testing for our exercise, the z-score helps determine the range or boundary of the interval. Additionally, it aids in comparing the calculated test statistic to a critical value from the standard normal distribution to decide whether or not to reject the null hypothesis.

The z-score is found using critical values from z-tables or statistical calculators for standard levels of confidence, such as 98% for confidence intervals. For hypothesis testing at a 1% significance level, the critical z is much larger, aiding in determining statistical significance.

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Most popular questions from this chapter

In parts of the eastern United States, whitctail deer are a major nuisance to farmers and homeowners, frequently damaging crops, gardens, and landscaping. A consumer organization arranges a test of two of the leading deer repellents \(\mathrm{A}\) and \(\mathrm{B}\) on the market. Fifty-six unfenced gardens in areas having high concentrations of deer are used for the test. Twenty-nine gardens are chosen at random to receive repellent \(\mathrm{A}\), and the other 27 receive repellent \(\mathrm{B}\). For each of the 56 gardens, the time elapsed between application of the repellent and the appearance in the garden of the first deer is recorded. For repellent \(\mathrm{A}\), the mean time is 101 hours. For repellent \(B\), the mean time is 92 hours. Assume that the two populations of elapsed times have normal distributions with population standard deviations of 15 and 10 hours, respectively. a. Let \(\mu_{1}\) and \(\mu_{2}\) be the population means of elapsed times for the two repellents, respectively. Find the point estimate of \(\mu_{1}-\mu_{2}\). b. Find a \(97 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) c. Test at a \(2 \%\) significance level whether the mean elapsed times for repellents \(A\) and \(B\) are different. Use both approaches, the critical-value and \(p\) -value, to perform this test.

A high school counselor wanted to know if tenth-graders at her high school tend to have more free time than the twelfth-graders. She took random samples of 25 tenth-graders and 23 twelfth-graders. Each student was asked to record the amount of free time he or she had in a typical week. The mean for the tenthgraders was found to be 29 hours of free time per week with a standard deviation of \(7.0\) hours. For the twelfthgraders, the mean was 22 hours of free time per week with a standard deviation of \(6.2\) hours. Assume that the two populations are normally distributed with equal but unknown population standard deviations. a. Make a \(90 \%\) confidence interval for the difference between the corresponding population means. b. Test at a \(5 \%\) significance level whether the two population means are different.

A company claims that its medicine, Brand A, provides faster relief from pain than another company's medicine, Brand B. A researcher tested both brands of medicine on two groups of randomly selected patients. The results of the test are given in the following table. The mean and standard deviation of relief times are in minutes. \begin{tabular}{cccc} \hline Brand & Sample Size & Mean of Relief Times & Standard Deviation of Relief Times \\ \hline A & 25 & 44 & 11 \\ B & 22 & 49 & 9 \\ \hline \end{tabular} Assume that the two populations are normally distributed with unknown but equal standard deviations. a. Construct a \(99 \%\) confidence interval for the difference between the mean relief times for the two brands of medicine. b. Test at a \(1 \%\) significance level whether the mean relief time for Brand \(\mathrm{A}\) is less than that for Brand B.

A November 2011 Gallup poll asked American adults about their views of healthcare and the healthcare system in the United States. Although feelings about the quality of healthcare were positive, the same cannot be said about the quality of the healthcare system. According to this study, \(29 \%\) of Independents and \(27 \%\) of Democrats rated the healthcare system as being excellent or good (www.gallup.com/poll/ \(150788 /\) Americans-Maintain- Negative-View-Healthcare-Coverage.aspx). Suppose that these results were based on samples of 1200 Independents and 1300 Democrats. a. Let \(p_{1}\) and \(p_{2}\) be the proportions of all Independents and all Democrats, respectively, who will rate the healthcare system as being excellent or good. Construct a \(97 \%\) confidence interval for \(p_{1}-p_{2}\) b. Using a \(1 \%\) significance level, can you conclude that \(p_{1}\) is different from \(p_{2}\) ? Use both the critical-value and the \(p\) -value approaches.

The following information is obtained from two independent samples selected from two populations. $$ \begin{array}{lll} n_{1}=650 & \bar{x}_{1}=1.05 & \sigma_{1}=5.22 \\ n_{2}=675 & \bar{x}_{2}=1.54 & \sigma_{2}=6.80 \end{array} $$ ah What is the point estimate of \(\mu_{1}-\mu_{2}\) ? b. Construct a 95\% confidence interval for \(\mu_{1}-\mu_{2}\). Find the margin of error for this estimate.

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