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Chapter 17: Problem 41

Athletes performing in bright sunlight often smear black eye grease under their eyes to reduce glare. Does eye grease work? In one study, 16 student subjects took a test of sensitivity to contrast after three hours facing into bright sun, both with and without eye grease. This is a matched pairs design. If eye grease is effective, subjects will be more sensitive to contrast when they use eye grease. Here are the differences in sensitivity, with eye grease minus without eye grease: We want to know whether eye grease increases sensitivity on the average. (a) What are the null and alternative hypotheses? Say in words what mean \(\mu\) your hypotheses concern. (b) Suppose that the subjects are an SRS of all young people with normal vision, that contrast differences follow a Normal distribution in this population, and that the standard deviation of differences is \(\sigma=0.22\). Carry out a test of significance.

Short Answer

Expert verified
Test \(H_0: \mu = 0\) against \(H_a: \mu > 0\) using a matched pairs t-test; reject \(H_0\) if p-value < \(0.05\).

Step by step solution

01

Define Hypotheses

Begin by stating the null and alternative hypotheses. The null hypothesis \(H_0\) is that the mean difference in sensitivity is zero, meaning eye grease does not affect sensitivity. The alternative hypothesis \(H_a\) is that the mean difference in sensitivity is greater than zero, meaning eye grease increases sensitivity. Let \(\mu\) represent the mean of the differences in sensitivity. Thus, the hypotheses are: \(H_0: \mu = 0\) and \(H_a: \mu > 0\).
02

Collect Sample Data

Obtain the data of sensitivity differences for the 16 subjects, denoted \(d_i\), where \(i = 1, 2, ..., 16\). Calculate the sample mean difference \(\bar{d}\) and the sample standard deviation \(s_d\).
03

Compute Test Statistic

Use the test statistic for a matched pairs t-test, given as \(t = \frac{\bar{d} - \mu}{s/\sqrt{n}}\). Here \(\bar{d}\) is the sample mean difference, \(\mu = 0\) for the null hypothesis, \(s\) is the sample standard deviation, and \(n = 16\) is the number of subjects.
04

Determine Critical Value or P-Value

Find the critical t-value for a one-tailed test with \(n-1 = 15\) degrees of freedom using a statistical table or software. Alternatively, compute the p-value corresponding to the calculated test statistic.
05

Make a Decision

Compare the p-value to the significance level (often \(\alpha = 0.05\)). If the p-value is less than \(\alpha\), reject the null hypothesis; otherwise, do not reject it.

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Key Concepts

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

Matched Pairs Design
A matched pairs design is an experimental technique where two sets of data are paired together based on certain criteria. In our example, the study aims to determine if eye grease improves sensitivity to contrast in bright sunlight. Each participant takes the test twice: once with and once without the eye grease. This pairing allows for a direct comparison of each individual's performance under both conditions.

This approach helps control for individual differences between subjects that could affect sensitivity. For example, some people might naturally have better contrast sensitivity than others, regardless of using eye grease. By using each subject as their own control, it reduces variation in the data that isn't due to the treatment itself.

Key aspects of matched pairs design include:
  • Using the same subjects for both conditions reduces variability.
  • It enhances the ability to detect differences attributed to the treatment.
  • Is especially useful in small samples like the 16 subjects in this study.
Null and Alternative Hypotheses
Before conducting any statistical test, it's crucial to establish the null and alternative hypotheses. These statements set the foundation for analyzing the data and drawing conclusions.

In the eye grease study, we define:

  • **Null Hypothesis ( $H_0$**): The mean difference in contrast sensitivity, $ar{d}$ , is equal to 0. This indicates that the eye grease does not have an effect.
  • **Alternative Hypothesis ( $H_a$**): The mean difference in contrast sensitivity is greater than 0, suggesting that the eye grease enhances sensitivity.
The null hypothesis always suggests no effect or difference, serving as the standard by which the experimental outcome is measured. If the analysis shows results substantially different from this hypothesis, it increases the likelihood that the alternative hypothesis better explains the data.

The symbol $ar{d}$ represents the sample mean difference, reflecting the average change in sensitivity with eye grease. Setting these hypotheses helps determine if observed differences are statistically significant.
t-Test
A t-test is a statistical method used to determine if there is a significant difference between the means of two groups. In our matched pairs study, a t-test compares sensitivity differences between conditions with and without eye grease.

Here's how the t-test works in this context:
  • Calculate the mean ( \(ar{d}\)) and standard deviation ( \(s_d\)) of the differences in contrast sensitivity.
  • Use these values in the formula: \(t = \frac{\bar{d} - 0}{s_d / \sqrt{n}}\), where \(n\) is the number of observations (16 subjects).
  • The null hypothesis mean is 0, indicating no effect from the eye grease.
Next, find the critical t-value from statistical tables or software, which depends on the degrees of freedom (number of subjects minus one).

Comparing the calculated t-statistic with this critical value determines whether the result is statistically significant. A significant result means we can reject the null hypothesis and conclude that eye grease likely improves contrast sensitivity in the sample. If not significant, the data does not provide enough evidence against the null hypothesis, suggesting any difference might be due to random chance.

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Most popular questions from this chapter

You use software to carry out a test of significance. The program tells you that the \(P\)-value is \(P=0.011\). You conclude that the probability, computed assuming that \(\mathrm{H}_{0}\) is (a) true, of the test statistic taking a value as extreme as or more extreme than that actually observed is \(0.011\). (b) true, of the test statistic taking a value as extreme as or less extreme than that actually observed is \(0.011\). (c) false, of the test statistic taking a value as extreme as or more extreme than that actually observed is \(0.011\).

The Graduate Management Admission Test (GMAT) is taken by individuals interested in pursuing graduate management education. GMAT scores are used as part of the admissions process for more than 6100 graduate management programs worldwide. The mean score for all test-takers is 550 with a standard deviation of \(120 .^{1}\) A researcher in the Philippines is concerned about the performance of undergraduates in the Philippines on the GMAT. She believes that the mean scores for this year's college seniors in the Philippines who are interested in pursuing graduate management education will be less than 550 . She has a random sample of 250 college seniors in the Philippines interested in pursuing graduate management education take the GMAT. Suppose we know that GMAT scores are Normally distributed with standard deviation \(\sigma=120\). (a) We seek evidence against the claim that \(\mu=550\). What is the sampling distribution of the mean score \(\mathrm{x}^{-} \bar{x}\) of a sample of 250 students if the claim is true? Draw the density curve of this distribution. (Sketch a Normal curve, then mark on the axis the values of the mean and 1,2 , and 3 standard deviations of the sampling distribution on either side of the mean.) (b) Suppose that the sample data give \(\mathrm{x}^{-} \bar{x}=542\). Mark this point on the axis of your sketch. (c) Suppose that the sample data give \(\mathrm{x}^{-} \bar{x}=532\). Mark this point on your sketch. Using your sketch, explain in simple language why one result is good evidence that the mean score of all college seniors in the Philippines interested in pursuing graduate management education who plan to take the GMAT is less than 550 and why the other outcome is not.

You use software to carry out a test of significance. The program tells you that the \(P\)-value is \(P=0.011\). This result is (a) not statistically significant at either \(\alpha=0.05\) or \(\alpha=0.01\). (b) statistically significant at \(\alpha=0.05\) but not at \(\alpha=0.01\). (c) statistically significant at both \(\alpha=0.05\) and \(\alpha=0.01\).

The \(z\) statistic for a one-sided test is \(z=2.29\). This test is (a) not statistically significant at either \(\alpha=0.05\) or \(\alpha=0.01\). (b) statistically significant at \(\alpha=0.05\) but not at \(\alpha=0.01\). (c) statistically significant at both \(\alpha=0.05\) and \(\alpha=0.01\).

Successful hotel managers must have personality characteristics traditionally stereotyped as feminine (such as "compassionate") as well as those traditionally stereotyped as masculine (such as "forceful"). The Bem Sex-Role Inventory (BSRI) is a personality test that gives separate ratings for "female" and "male" stereotypes, both on a scale of 1 to 7 . Although the BSRI was developed at a time when these stereotypes were more pronounced, it is still widely used to assess personality types. Unfortunately, the ratings are often referred to as femininity and masculinity scores. A sample of 148 male general managers of three-star and four-star hotels had mean BSRI femininity score \(\mathrm{x}^{-} \bar{x}=5.29 .{ }^{6}\) The mean score for the general male population is \(\mu=5.19\). Do hotel managers, on the average, differ statistically significantly in femininity score from men in general? Assume that the standard deviation of scores in the population of all male hotel managers is the same as the \(\sigma=0.78\) for the adult male population. (a) State null and alternative hypotheses in terms of the mean femininity score \(\mu\) for male hotel managers. (b) Find the \(z\) test statistic. (c) What is the \(P\)-value for your \(z\) ? What do you conclude about male hotel managers?
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