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Chapter 10: Problem 29

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.

Short Answer

Expert verified
a. The 99% confidence interval for the difference in means is calculated to be [..., ...]. b. The calculated test statistic is ..., which falls ... the critical value of -2.33. Therefore, at the 1% significance level, there is/is not sufficient evidence to say that Brand A has a lower mean relief time than Brand B.

Step by step solution

01

Determine the sample statistics

Identify the relevant statistics from the given data: for Brand A, the sample size (n1) is 25, the sample mean (x1_bar) is 44 minutes, and the sample standard deviation (s1) is 11 minutes; for Brand B, the sample size (n2) is 22, the sample mean (x2_bar) is 49 minutes, and the sample standard deviation (s2) is 9 minutes.
02

Calculate pooled standard deviation

Find the pooled standard deviation, Sp, using the formula: \[Sp = \sqrt{((n1-1)*s1^2 + (n2-1)*s2^2)/(n1+n2-2)}\]
03

Determine the standard error of the difference in means

Calculate the standard error of the difference in means, SE_diff, using the formula: \[SE_diff = Sp*\sqrt{(1/n1) + (1/n2)}\]
04

Calculate 99% confidence interval for difference in means

Using the z-score for a 99% confidence interval (approximately 2.58), calculate the confidence interval using the formula: \[(x1_bar - x2_bar) \pm z*SE_diff\]
05

Set up and carry out the hypothesis test

Conduct a one-tailed hypothesis test at a significance level of 1%. The null hypothesis is that the means are equal, while the alternative is that the mean of Brand A is less than that of Brand B. Use the calculated SE_diff to find the test statistic: \[z = (x1_bar - x2_bar)/SE_diff\] Then, compare this to the critical value at the specified significance level (approximately -2.33 for a one-tailed test at 1%) to make the decision.

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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 gives us a range of values within which we believe the true difference in means lies. This range is defined by a lower limit and an upper limit. For instance, if we are calculating a 99% confidence interval, like in the exercise, we are 99% sure that the true difference between the mean relief time of Brand A and Brand B falls within this range.

To calculate the confidence interval for the difference in means, we use the formula: \[(x1\_bar - x2\_bar) \pm z*SE\_diff\]Here,
  • \(x1\_bar\) and \(x2\_bar\) are the sample means for Brand A and Brand B respectively,
  • \(z\) is the z-score corresponding to the 99% confidence level (approximately 2.58),
  • \(SE\_diff\) is the standard error of the difference in means.
The confidence interval is crucial because it helps determine if the observed difference in means is statistically significant or could have happened by chance. A wider interval indicates more variability in the data, whereas a narrower interval indicates less variability.
Pooled Standard Deviation
Pooled standard deviation is a way to estimate the common standard deviation of two groups where we assume that the true standard deviations are equal. When comparing two samples, this is especially useful as it gives a more accurate measure of combined variability.

The formula for calculating the pooled standard deviation \(Sp\) is given by: \[Sp = \sqrt{\frac{((n1-1)*s1^2 + (n2-1)*s2^2)}{n1+n2-2}} \]Where:
  • \(n1\) and \(n2\) are the sample sizes of Brand A and Brand B respectively,
  • \(s1\) and \(s2\) are the standard deviations of Brand A and Brand B respectively.
This calculated value, \(Sp\), combines the variability in both samples under the assumption of equal variances. This pooled method is particularly helpful for generating more reliable results when sample sizes vary between the two groups.
Standard Error of the Difference
The standard error of the difference (\(SE\_diff\)) quantifies the variability or standard deviation of the difference in sample means. It helps in determining how much the observed sample means differ by chance.

The formula to calculate this is:\[SE\_diff = Sp*\sqrt{(1/n1) + (1/n2)}\]Where:
  • \(Sp\) is the pooled standard deviation,
  • \(n1\) and \(n2\) are the sample sizes of Brand A and Brand B respectively.
Understanding the standard error of the difference is vital for both hypothesis testing and constructing confidence intervals. A smaller standard error suggests that the sample means are a reliable reflection of the population means, whereas a larger standard error might indicate more variability in the difference.
Z-Score
The z-score is a statistic that helps understand how far an observed sample mean is from the null hypothesis mean, measured in standard deviations. In hypothesis testing, it is compared to a critical value to determine if the findings are significant.

In our exercise, the z-score for the difference of means is calculated using:\[z = \frac{(x1\_bar - x2\_bar)}{SE\_diff} \]Where:
  • \(x1\_bar\) and \(x2\_bar\) are the sample means of Brand A and Brand B,
  • \(SE\_diff\) is the standard error of the difference in means.
The z-score tells us how many standard errors the sample difference is from zero (the null hypothesis difference). If the z-score falls beyond the critical value in a hypothesis test (like -2.33 for a one-tailed test at 1% significance), we reject the null hypothesis and conclude there is a significant difference between the means.

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Most popular questions from this chapter

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 a \(1 \%\) significance level, can you conclude that the proportions of all men and all women who play the lottery often are different?

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 a \(1 \%\) 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.

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 \(\bar{x}_{1} \geq \bar{x}_{2}\) ?

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.

A mail-order company has two warehouses, one on the West Coast and the second on the East Coast. The company's policy is to mail all orders placed with it within 72 hours. The company's quality control department checks quite often whether or not this policy is maintained at the two warehouses. A recently taken sample of 400 orders placed with the warehouse on the West Coast showed that 364 of them were mailed within 72 hours. Another sample of 300 orders placed with the warehouse on the East Coast showed that 279 of them were mailed within 72 hours. a. Construct a \(97 \%\) confidence interval for the difference between the proportions of all orders placed at the two warehouses that are mailed within 72 hours. b. Using a \(2.5 \%\) significance level, can you conclude that the proportion of all orders placed at the warehouse on the West Coast that are mailed within 72 hours is lower than the corresponding proportion for the warehouse on the East Coast?
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