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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. $$ \begin{array}{cccc} \hline & & \text { Mean of } & \text { Standard Deviation } \\ \text { Brand } & \text { Sample Size } & \text { Relief Times } & \text { of Relief Times } \\ \hline \text { A } & 25 & 44 & 11 \\ \text { B } & 22 & 49 & 9 \\ \hline \end{array} $$ a. Construct a \(99 \%\) confidence interval for the difference between the mean relief times for the two brands of medicine. b. Test at the \(1 \%\) 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 between the mean relief times for the two brands of medicine is calculated as discussed in the solution steps. b. If the calculated Z-Score in the significance test is less than the critical Z-Score, we can conclude that the mean relief time for Brand A is less than that for Brand B at a 1% significance level.

Step by step solution

01

Compute the Difference in Sample Means

Calculate the difference between the two sample means, which is \(\bar{x}_A - \bar{x}_B = 44 - 49 = -5\). So, the sample mean difference is -5 minutes.
02

Compute the Standard Error

We calculate the standard error, which is given by \(\sqrt{\frac{s_A^2}{n_A} + \frac{s_B^2}{n_B}} = \sqrt{\frac{11^2}{25} + \frac{9^2}{22}} \). This gives us the standard error, which measures the accuracy of our sample difference.
03

Compute the Confidence Interval

From the Normal Distribution table, we find that the Z-Score for a 99% confidence interval is 2.58. We calculate the lower and upper limit as: Lower Limit = \(\bar{x}_{A-B} - Z*SE\) and Upper Limit = \(\bar{x}_{A-B} + Z*SE\), which gives us the 99% confidence interval for the difference of means.
04

Formulate Hypotheses for the Significance Test

Formulate the null hypothesis(H0) as the mean relief time for Brand A equals that of Brand B, and the alternative hypothesis(HA) as the mean relief time for Brand A is less than that for Brand B.
05

Conduct the Significance Test

To test the null hypothesis, we calculate the Z-Score as: \(Z = \frac{\bar{x}_{A-B} - \mu_0}{SE}\). The null hypothesis is rejected if the calculated Z-Score is less than the critical Z-Score for a 1% significance level.

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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 estimates the true difference in population means. It's like a safety net telling us how confident we are about these estimates.
For a 99% confidence interval, we use a Z-Score of 2.58, meaning we are 99% sure the true mean difference falls within this range.
Here's how to calculate it:
  • Find the difference in sample means: \(-5\) minutes for Brand A compared to Brand B.
  • Calculate the standard error to measure the variation between sample means.
  • Determine the interval by adding and subtracting the product of Z-Score and standard error from the mean difference.
This interval helps us understand if Brand A really offers faster relief, and if the guess isn't just due to random chance.
Hypothesis Testing
Hypothesis testing is the process of making decisions or inferences about population parameters based on sample data. Think of it as testing a claim with evidence.
In this exercise, we compare two hypotheses:
  • Null Hypothesis (\(H_0\)): Brand A's mean relief time is equal to Brand B's.
  • Alternative Hypothesis (\(H_A\)): Brand A's mean relief time is less than Brand B's.
To make this decision, we look at the test statistic (Z-Score) and compare it to a critical value. If it falls in the rejection region, we reject the null hypothesis. Here, we're using a 1% significance level, which means we want strong evidence before making a judgment.
Ultimately, hypothesis testing helps in determining whether the results about relief times are statistically significant.
Z-Score
The Z-Score is a statistical measure that explains how far away our sample mean is from the population mean in terms of standard deviations. It's a way to understand where our sample lies on a normal distribution curve.
In this problem, we use it both for constructing the confidence interval and for hypothesis testing.
  • For the 99% confidence interval, a Z-Score of 2.58 is used to account for almost the entire bell curve while calculating the range.
  • For hypothesis testing, the Z-Score helps determine whether the observed difference between Brand A and Brand B is significant enough to reject the null hypothesis.
It's what helps bridge the gap between sample data and population insights.
Standard Error
Standard error measures how much your sample mean is expected to vary from the true population mean. Imagine it as a way of quantifying the precision of your sample statistics.
To compute it, you're looking at the variability in your data and considering the sample size. For this exercise:
  • We calculate standard error using the formula: \[ \text{SE} = \sqrt{\frac{s_A^2}{n_A} + \frac{s_B^2}{n_B}} \]
  • This gives us an overall understanding of the spread of sample means around the true population mean.
The smaller the standard error, the more reliable our sample mean difference is, allowing us to make better inferences about the overall population.

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Most popular questions from this chapter

The manager of a factory has devised a detailed plan for evacuating the building as quickly as possible in the event of a fire or other emergency. An industrial psychologist believes that workers actually leave the factory faster at closing time without following any system. The company holds fire drills periodically in which a bell sounds and workers leave the building according to the system. The evacuation time for each drill is recorded. For comparison, the psychologist also records the evacuation time when the bell sounds for closing time each day. A random sample of 36 fire drills showed a mean evacuation time of \(5.1\) minutes with a standard deviation of \(1.1\) minutes. A random sample of 37 days at closing time showed a mean evacuation time of \(4.2\) minutes with a standard deviation of \(1.0\) minute. a. Construct a \(99 \%\) confidence interval for the difference between the two population means. b. Test at the \(5 \%\) significance level whether the mean evacuation time is smaller at closing time than during fire drills.

Several retired bicycle racers are coaching a large group of young prospects. They randomly select seven of their riders to take part in a test of the effectiveness of a new dietary supplement that is supposed to increase strength and stamina. Each of the seven riders does a time trial on the same course. Then they all take the dietary supplement for 4 weeks. All other aspects of their training program remain as they were prior to the time trial. At the end of the 4 weeks, these riders do another time trial on the same course. The times (in minutes) recorded by each rider for these trials before and after the 4-week period are shown in the following table. $$ \begin{array}{l|lllrrll} \hline \text { Before } & 103 & 97 & 111 & 95 & 102 & 96 & 108 \\ \hline \text { After } & 100 & 95 & 104 & 101 & 96 & 91 & 101 \\ \hline \end{array} $$ a. Construct a \(99 \%\) confidence interval for the mean \(\mu_{d}\) of the population paired differences, where a paired difference is equal to the time taken before the dietary supplement minus the time taken after the dietary supplement. b. Test at the \(2.5 \%\) significance level whether taking this dietary supplement results in faster times in the time trials.

As was mentioned in Exercise \(9.38\), The Bath Heritage Days, which take place in Bath, Maine, switched to a Whoopie Pie eating contest in 2009 . Suppose the contest involves eating nine Whoopie Pies, each weighing \(1 / 3\) pound. The following data represent the times (in seconds) taken by each of the 13 contestants (all of whom finished all nine Whoopie Pies) to eat the first Whoopie Pie and the last (ninth) Whoopie Pie. $$ \begin{array}{l|rrrrrrrrrrrr} \hline \text { Contestant } & \mathbf{1} & \mathbf{2} & \mathbf{3} & \mathbf{4} & \mathbf{5} & \mathbf{6} & \mathbf{7} & \mathbf{8} & \mathbf{9} & \mathbf{1 0} & \mathbf{1 1} & \mathbf{1 2} & \mathbf{1 3} \\ \hline \text { First pie } & 49 & 59 & 66 & 49 & 63 & 70 & 77 & 59 & 64 & 69 & 60 & 58 & 71 \\ \hline \text { Last pie } & 49 & 74 & 92 & 93 & 91 & 73 & 103 & 59 & 85 & 94 & 84 & 87 & 111 \\ \hline \end{array} $$ 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 eat the first pie. b. Using the \(10 \%\) 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.

Sixty-five percent of all male voters and \(40 \%\) of all female voters favor a particular candidate. A sample of 100 male voters and another sample of 100 female voters will be polled. What is the probability that at least 10 more male voters than female voters will favor this candidate?

A sample of 500 observations taken from the first population gave \(x_{1}=305\). Another sample of 600 observations taken from the second population gave \(x_{2}=348\). a. Find the point estimate of \(p_{1}-p_{2}\). b. Make a \(97 \%\) confidence interval for \(p_{1}-p_{2}\). c. Show the rejection and nonrejection regions on the sampling distribution of \(\hat{p}_{1}-\hat{p}_{2}\) for \(H_{0}: p_{1}=p_{2}\) versus \(H_{1}: p_{1}>p_{2} .\) Use a significance level of \(2.5 \% .\) d. Find the value of the test statistic \(z\) for the test of part \(\mathrm{c}\). e. Will you reject the null hypothesis mentioned in part \(\mathrm{c}\) at a significance level of \(2.5 \%\) ?

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