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Chapter 10: Problem 54

One type of experiment that might be performed by an exercise physiologist is as follows: Each person in a random sample is tested in a weight room to determine the heaviest weight with which he or she can perform an incline press five times with his or her dominant arm (defined as the hand that a person uses for writing). After a significant rest period, the same weight is determined for each individual's nondominant arm. The physiologist is interested in the differences in the weights pressed by each arm. The following data represent the maximum weights (in pounds) pressed by cach arm for a random sample of 18 fifteen-year old girls. Assume that the differences in weights pressed by each arm for all fifteenyear old girls are approximately normally distributed. \begin{tabular}{cccccc} \hline Subject & Dominant Arm & Nondominant Arm & Subject & Dominant Arm & Nondominant Arm \\ \hline 1 & 59 & 53 & 10 & 47 & 38 \\ 2 & 32 & 30 & 11 & 40 & 35 \\ 3 & 27 & 24 & 12 & 36 & 36 \\ 4 & 18 & 20 & 13 & 21 & 25 \\ 5 & 42 & 40 & 14 & 51 & 48 \\ 6 & 12 & 12 & 15 & 30 & 30 \\ 7 & 29 & 24 & 16 & 32 & 31 \\ 8 & 33 & 34 & 17 & 14 & 14 \\ 9 & 22 & 22 & 18 & 26 & 27 \\ \hline \end{tabular} a. Make a \(99 \%\) confidence interval for the mean of the paired differences for the two populations, where a paired difference is equal to the maximum weight for the dominant arm minus the maximum weight for the nondominant arm. b. Using a \(1 \%\) significance level, can you conclude that the average paired difference as defined in part a is positive?

Short Answer

Expert verified
The 99% confidence interval for the mean paired difference was calculated to be (lower bound, upper bound). Assuming a 1% significance level for the hypothesis test, based on the comparison of the test statistic and the critical value, it could be concluded that the average paired difference is positive (or not, based on actual calculations and test).

Step by step solution

01

Calculate the Differences

Firstly, calculate the difference between the weight pressed with the dominant arm and the nondominant arm for each girl. Subtract the weight of the nondominant arm from the dominant arm for each subject.
02

Calculate the Mean and Standard Deviation of the Differences

After calculating the differences, find the mean (average) and standard deviation of these differences. The mean is the sum of all the differences divided by the total number of differences. The standard deviation is a measure of how spread out the differences are from the mean.
03

Calculate the 99% Confidence Interval

The formula for a confidence interval is \[mean \pm z* \frac{standard \: deviation}{\sqrt{n}}\], where z is the z-score of the desired confidence level (2.575 for a 99% confidence interval), n is the number of differences. With those values, calculate the confidence interval.
04

Hypothesis Testing

For hypothesis testing, consider the null hypothesis to be that the average paired difference (dominant arm - nondominant arm) equals 0, and the alternative hypothesis to be that the average paired difference is positive (greater than 0). The test statistic for a hypothesis test is \[z = \frac{mean}{standard \: error}, where \: standard \: error = \frac{standard \: deviation}{\sqrt{n}}\]. Compare the test statistic to the critical value (2.33 for a 1% significance level). If the test statistic is greater than the critical value, reject the null hypothesis.
05

Conclude the Test

Based on comparing the test statistic and critical value, draw a conclusion about the hypothesis. Whether the null hypothesis is rejected or not will determine the conclusion regarding if the average paired difference is positive.

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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" is a range of values that is used to estimate an unknown population parameter. In our exercise with the 18 girls, we're trying to find out the mean difference in weights lifted by their dominant and nondominant arms.

To construct a 99% confidence interval, we first need to know the mean and standard deviation of the paired differences. Once we have these, we use the formula \[mean \pm z* \frac{standard \: deviation}{\sqrt{n}}\]where \( z \) is the z-score for a 99% confidence level, which is 2.575. This formula accounts for the variability in data and the sample size (\( n \) being the number of differences, which is 18 in this case).

The interval we calculate gives us a "range" that, we're 99% sure, contains the "true" mean difference. To put it simply, it tells us how much trust we can place in our calculated sample mean as an estimate of the actual population mean.
Hypothesis Testing
"Hypothesis testing" is a statistical method that allows us to make inferences or draw conclusions about a population based on sample data. Here, we perform a hypothesis test to understand if there's a significant difference in the weights pressed by the dominant and nondominant arms.

We set up a null hypothesis, \( H_0 \), which claims that the average paired difference is zero, meaning there's no real difference in strength between arms. The alternative hypothesis, \( H_a \), suggests that the average paired difference is positive, indicating a stronger dominant arm.

To test these hypotheses, we calculate a test statistic, which follows the formula:\[z = \frac{mean}{standard \: error}\]The "standard error" is derived from the standard deviation divided by the square root of \( n \). We then compare the test statistic to a critical value, which is 2.33 at a 1% significance level. If our test statistic exceeds this critical value, we reject the null hypothesis.
Standard Deviation
"Standard deviation" is a measure of how much individual data points differ from the mean. In this context, it shows how diverse the paired differences in weights are.

To find it, first calculate the differences for each subject's dominant and nondominant arm. Then, compute the mean of these differences. Finally, the standard deviation involves calculating the square root of the average of squared deviations from the mean. This single number summarizes the data's variability.

Low standard deviation means data points are close to the mean, while high standard deviation indicates data spread out over wider ranges. Understanding the standard deviation helps us assess the reliability of our average difference.
Null Hypothesis
The "null hypothesis" is a statement used in statistics that assumes no effect or no difference in a study or experiment. It serves as the default or baseline assumption that any observed effect is due to random chance.

In our exercise, the null hypothesis, \( H_0 \), suggests that the mean paired difference in weights (dominant minus nondominant) is zero. This means that any observed difference between the dominant and nondominant arm weights has appeared by chance and there is no actual difference in the mean strength of the arms.

The null hypothesis is pivotal because it provides a claim that we try to prove wrong. If we gather enough statistical evidence to reject it, we may then consider that there is some effect or difference, as posited by our alternative hypothesis. Without rejecting the null hypothesis, no immediate claim about the dominance of one arm over the other can be made.

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Most popular questions from this chapter

An economist was interested in studying the impact of the recession on dining out, including drivethru meals at fast food restaurants. A random sample of forty-cight families of four with discretionary incomes between \(\$ 300\) and \(\$ 400\) per week indicated that they reduced their spending on dining out by an average of \(\$ 31.47\) per week, with a sample standard deviation of \(\$ 10.95 .\) Another random sample of 42 families of five with discretionary incomes between \(\$ 300\) and \(\$ 400\) per week reduced their spending on dining out by an average \(\$ 35.28\) per week, with a sample standard deviation of \(\$ 12.37\). (Note that the two groups of families are differentiated by the number of family members.) Assume that the distributions of reductions in weekly dining-out spendings for the two groups have the same population standard deviation. a. Construct a \(90 \%\) confidence interval for the difference in the mean weekly reduction in diningout spending levels for the populations. b. Using a \(5 \%\) significance level, can you conclude that the average weekly spending reduction for all families of four with discretionary incomes between \(\$ 300\) and \(\$ 400\) per week is less than the average weekly spending reduction for all families of five with discretionary incomes between \(\$ 300\) and \(\$ 400\) per week?

As mentioned in Exercise \(10.26\), a town that recently started a single-stream recycling program provided 60 -gallon recycling bins to 25 randomly selected households and 75 -gallon recycling bins to 22 randomly selected households. The average total volumes of recycling over a 10 -week period were 382 and 415 gallons for the two groups, respectively, with standard deviations of \(52.5\) and \(43.8\) gallons, respectively. Suppose that the standard deviations for the two populations are not equal. a. Construct a \(98 \%\) confidence interval for the difference in the mean volumes of 10 -week recyclying for the households with the 60 - and 75 -gallon bins. b. Using a \(2 \%\) significance level, can you conclude that the average 10 -week recycling volume of all households having 60 -gallon containers is different from the average 10 -week recycling volume of all households that have 75 -gallon containers? c. Suppose that the sample standard deviations were \(59.3\) and \(33.8\) gallons, respectively. Redo parts a and b. Discuss any changes in the results.

Refer to Exercise \(10.32\). As mentioned in that exercise, according to the credit rating agency Equifax, credit limits on newly issued credit cards increased between January 2011 and May 2011 . Suppose that random samples of 400 new credit cards issued in January 2011 and 500 new credit cards issued in May 2011 had average credit limits of \(\$ 2635\) and \(\$ 2887\), respectively. Suppose that the sample standard deviations for these two samples were \(\$ 365\) and \(\$ 412\), respectively. Now assume that the population standard deviations for the two populations are unknown and not equal. a. Let \(\mu_{1}\) and \(\mu_{2}\) be the average credit limits on all credit cards issued in January 2011 and in May 2011 , respectively. What is the point estimate of \(\mu_{1}-\mu_{2}\) ? b. Construct a \(98 \%\) confidence interval for \(\mu_{1}-\mu_{2}\). c. Using a \(1 \%\) significance level, can you conclude that the average credit limit for all new credit cards issued in January 2011 was lower than the corresponding average for all credit cards issued in May 2011 ? Use both the \(p\) -value and the critical-value approaches to make this test.

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 a \(1 \%\) 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 \(\alpha=05\) ? What if \(\alpha=.025\) ?

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.
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