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Do Hands Adapt to Water? Researchers in the UK designed a study to determine if skin wrinkled from submersion in water performed better at handling wet objects. \(^{62}\) They gathered 20 participants and had each move a set of wet objects and a set of dry objects before and after submerging their hands in water for 30 minutes (order of trials was randomized). The response is the time (seconds) it took to move the specific set of objects with wrinkled hands minus the time with unwrinkled hands. The mean difference for moving dry objects was 0.85 seconds with a standard deviation of 11.5 seconds. The mean difference for moving wet objects was -15.1 seconds with a standard deviation of 13.4 seconds. (a) Perform the appropriate test to determine if the wrinkled hands were significantly faster than unwrinkled hands at moving dry objects. (b) Perform the appropriate test to determine if the wrinkled hands were significantly faster than unwrinkled hands at moving wet objects.

Short Answer

Expert verified
For the dry objects, if the calculated t is greater than the critical value, reject the null hypothesis which means the wrinkled hands are significantly faster. For the wet objects, if the calculated t is less than critical value, reject the null hypothesis which means the wrinkled hands are significantly faster.

Step by step solution

01

Define the Hypotheses for Part (a)

For the first part, the null hypothesis \(H_{0}: \mu_{D} = 0\), where \(\mu_{D}\) is the mean difference for moving dry objects. The alternative hypothesis \(H_{1}: \mu_{D} > 0\).
02

Compute the Test Statistic for Part (a)

For the test statistic, calculate \(t = \frac{\bar{X}-\mu_{0}}{s / \sqrt{n}}\), where \(\bar{X}\) is the sample mean (0.85), \(\mu_{0}\) is the null hypothesis mean (0), s is the standard deviation (11.5), and n is the sample size (20).
03

Determine the Critical Value and Make a Decision for Part (a)

The critical value determined from t-distribution table with \(df = n - 1\) (19) at the 5% level of significance for a one-tailed test. If computed t is greater than the critical value, reject the null hypothesis.
04

Define the Hypotheses for Part (b)

For the second part, the null hypothesis \(H_{0}: \mu_{W} = 0\), where \(\mu_{W}\) is the mean difference for moving wet objects. The alternative hypothesis \(H_{1}: \mu_{W} < 0\).
05

Compute the Test Statistic for Part (b)

The test statistic is calculated in the same way but using the sample mean (-15.1), the standard deviation (13.4), and the sample size (20).
06

Determine the Critical Value and Make a Decision for Part (b)

The critical value is determined in the same way as in step 3 but the t-value is compared for the left tail of the distribution. If the calculated t is less than the critical value, reject the null hypothesis.

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Key Concepts

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

Null and Alternative Hypotheses
Understanding the null and alternative hypotheses is fundamental to hypothesis testing. In the context of our exercise, hypothesis testing seeks to determine whether the observed effects of wrinkled hands on moving objects are due to chance or a real effect. The null hypothesis (\( H_0 \)) represents a statement of no effect or no difference, essentially a baseline or default assumption. In part (a) of our exercise, the null hypothesis is that the mean difference in time for moving dry objects with wrinkled versus unwrinkled hands is zero (\( \mu_D = 0 \) ).

The alternative hypothesis (\( H_1 \)), on the other hand, represents what the researchers really want to test; it is the statement of an effect or difference. In the case of moving dry objects, the alternative hypothesis is that the mean difference is greater than zero (\( \mu_D > 0 \) ), indicating that wrinkled hands were faster. Similarly, for part (b) the alternative hypothesis postulates that the mean difference in time for moving wet objects with wrinkled hands is less than zero (\( \mu_W < 0 \) ), suggesting that wrinkled hands could be significantly faster at this task.

These hypotheses set the stage for the entire test, guiding the calculations and interpretations that follow.
Test Statistic Calculation
The test statistic is pivotal to hypothesis testing as it helps us decide whether to reject the null hypothesis. It is a standardized value that measures the degree of deviation from the null hypothesis. To compute the test statistic in part (a) for our exercise, we use the formula \( t = \frac{\bar{X}-\mu_{0}}{s / \sqrt{n}} \) where \(\bar{X}\) is the sample mean, \(\mu_{0}\) is the hypothesized population mean under the null, \(s\) is the sample standard deviation, and \(n\) is the sample size.

For part (a), \(\bar{X}\) is 0.85 seconds. The test statistic measures how many standard errors the sample mean is away from the null hypothesis mean. It considers both the size of the effect (the sample mean difference) and the precision of the estimate (the sample size and standard deviation). Plugging in the values, we get a specific test statistic that we can compare against a critical value to make a statistical decision.
t-Distribution Critical Value
The t-distribution critical value acts as a threshold in hypothesis testing. To determine whether our test statistic is extreme enough to reject the null hypothesis, we must compare it to a critical value from the t-distribution. This critical value depends on the desired level of significance (commonly 5%) and the degrees of freedom, which in turn depend on the sample size.

In the exercise, the degrees of freedom (df) for each test is 19, since it's calculated as the sample size minus one (\( n - 1 \)). Once we know the df, we can look up the critical value in a t-distribution table or use statistical software for more precise results. At a 5% significance level, the critical value is the point in the t-distribution where there's only a 5% chance of observing a more extreme value if the null hypothesis is true. The decision rule is straightforward: if the test statistic exceeds the critical value in a one-tailed test, we reject the null hypothesis.
One-tailed Test Significance
Finally, the concept of one-tailed test significance refers to the direction of the hypothesis. In a one-tailed test, the alternative hypothesis states that the parameter of interest is either strictly greater than or less than the null hypothesis value. It implies that researchers have a specific direction in mind for the effect being tested.

In our exercise, part (a) is a right-tailed test because we are testing if wrinkled hands are faster (a positive effect) at moving dry objects, so we are looking for a test statistic that exceeds the critical value on the right side of the t-distribution. Conversely, part (b) is a left-tailed test where we check if wrinkled hands are faster at moving wet objects, which corresponds to a negative mean difference (a negative effect), and we seek a test statistic that falls below the critical value on the left side. One-tailed tests are powerful when a specific direction is hypothesized because they can detect an effect in that direction with less data, increasing the test’s sensitivity.

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Most popular questions from this chapter

Find a \(95 \%\) confidence interval for the mean two ways: using StatKey or other technology and percentiles from a bootstrap distribution, and using the t-distribution and the formula for standard error. Compare the results. Mean price of a used Mustang car online, in \$1000s, using data in MustangPrice with \(\bar{x}=15.98\), \(s=11.11,\) and \(n=25\)

Use the t-distribution to find a confidence interval for a difference in means \(\mu_{1}-\mu_{2}\) given the relevant sample results. Give the best estimate for \(\mu_{1}-\mu_{2},\) the margin of error, and the confidence interval. Assume the results come from random samples from populations that are approximately normally distributed. A \(99 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) using the sample results \(\bar{x}_{1}=501, s_{1}=115, n_{1}=400\) and \(\bar{x}_{2}=469, s_{2}=96, n_{2}=200 .\)

Split the Bill? Exercise 2.153 on page 105 describes a study to compare the cost of restaurant meals when people pay individually versus splitting the bill as a group. In the experiment half of the people were told they would each be responsible for individual meal costs and the other half were told the cost would be split equally among the six people at the table. The data in SplitBill includes the cost of what each person ordered (in Israeli shekels) and the payment method (Individual or Split). Some summary statistics are provided in Table 6.20 and both distributions are reasonably bell-shaped. Use this information to test (at a \(5 \%\) level ) if there is evidence that the mean cost is higher when people split the bill. You may have done this test using randomizations in Exercise 4.118 on page 302 .

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Who Watches More TV: Males or Females? The dataset StudentSurvey has information from males and females on the number of hours spent watching television in a typical week. Computer output of descriptive statistics for the number of hours spent watching TV, broken down by gender, is given: \(\begin{array}{l}\text { Descriptive Statistics: TV } \\ \text { Variable } & \text { Gender } & \mathrm{N} & \text { Mean } & \text { StDev } \\ \text { TV } & \mathrm{F} & 169 & 5.237 & 4.100 \\ & \mathrm{M} & 192 & 7.620 & 6.427 \\\ \text { Minimum } & \mathrm{Q} 1 & \text { Median } & \mathrm{Q} 3 & \text { Maximum } \\ & 0.000 & 2.500 & 4.000 & 6.000 & 20.000 \\ & 0.000 & 3.000 & 5.000 & 10.000 & 40.000\end{array}\) (a) In the sample, which group watches more TV, on average? By how much? (b) Use the summary statistics to compute a \(99 \%\) confidence interval for the difference in mean number of hours spent watching TV. Be sure to define any parameters you are estimating. (c) Compare the answer from part (c) to the confidence interval given in the following computer output for the same data: \(\begin{array}{l}\text { Two-sample T for TV } \\ \text { Gender } & \text { N } & \text { Mean } & \text { StDev } & \text { SE Mean } \\ \mathrm{F} & 169 & 5.24 & 4.10 & 0.32 \\ \mathrm{M} & 192 & 7.62 & 6.43 & 0.46 \\ \text { Difference } & =\mathrm{mu}(\mathrm{F})-\mathrm{mu}(\mathrm{M}) & & \end{array}\) Estimate for for difference: -2.383 r difference: (-3.836,-0.930) \(99 \% \mathrm{Cl}\) for (d) Interpret the confidence interval in context.

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