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A random sample of 37 second graders who participated in sports had manual dexterity scores with mean 32.19 and standard deviation \(4.34 .\) An independent sample of 37 second graders who did not participate in sports had manual dexterity scores with mean 31.68 and standard deviation 4.56. a. Test to see whether sufficient evidence exists to indicate that second graders who participate in sports have a higher mean dexterity score. Use \(\alpha=.05\). b. For the rejection region used in part (a), calculate \(\beta\) when \(\mu_{1}-\mu_{2}=3\).

Short Answer

Expert verified
No evidence that sports improve scores. Power of test is 90.35% when \(\mu_1-\mu_2 = 3\).

Step by step solution

01

Define Hypotheses

State the null and alternative hypotheses for the test. The null hypothesis is that there is no difference in the mean dexterity scores between groups, i.e., \( H_0: \mu_1 = \mu_2 \). The alternative hypothesis is that the mean dexterity score for participants is higher, \( H_a: \mu_1 > \mu_2 \). This is a one-tailed test.
02

Determine Test Statistic

Since the sample sizes are equal and the samples are independent, we use the two-sample t-test formula for the test statistic:\[ t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}\]where \(\bar{x}_1 = 32.19\), \(\bar{x}_2 = 31.68\), \(s_1 = 4.34\), \(s_2 = 4.56\), and \(n_1 = n_2 = 37\). Substitute these values to calculate \(t\).
03

Calculate the t-value

Substitute the given values into the t-test formula:\[ t = \frac{32.19 - 31.68}{\sqrt{\frac{4.34^2}{37} + \frac{4.56^2}{37}}} = \frac{0.51}{\sqrt{\frac{18.8356}{37} + \frac{20.7936}{37}}} = \frac{0.51}{1.048} \approx 0.486\]
04

Determine Critical Value and Decision Rule

Using a t-distribution table, find the critical t-value for a significance level of \(\alpha = 0.05\) and \(df = 72\) (degrees of freedom using \(n_1 + n_2 - 2\)). The critical t-value is approximately 1.667. Reject \(H_0\) if the calculated t-value is greater than 1.667.
05

Conclusion for Part (a)

Since the calculated t-value of 0.486 is less than the critical t-value of 1.667, we do not reject the null hypothesis. There is not enough evidence at \(\alpha = 0.05\) to support the claim that second graders who participate in sports have a higher mean dexterity score.
06

Calculate Power of the Test for Part (b)

The power of a test is \(1 - \beta\), where \(\beta\) is the probability of Type II error. First, calculate the non-centrality parameter \(\delta\) when \(\mu_1 - \mu_2 = 3\):\[ \delta = \frac{\mu_1 - \mu_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} = \frac{3}{1.048} \approx 2.863\]
07

Determine Beta for Part (b)

Using the t-distribution, determine \(\beta\) by finding the probability that the t-statistic is less than the critical value (1.667) under the distribution with non-centrality parameter \(\delta = 2.863\). This results in \(\beta \approx 0.0965\).
08

Conclusion for Part (b)

The probability of error \(\beta\) when \(\mu_1 - \mu_2 = 3\) is approximately 0.0965, which implies the power of the test is \(1 - \beta \approx 0.9035\).

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Key Concepts

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

Understanding the T-Test
A t-test is a statistical method used to determine if there is a significant difference between the means of two groups. It helps us decide whether to accept or reject the null hypothesis. In the context of this exercise, we use a two-sample t-test because we are comparing the mean dexterity scores of two independent groups of second graders—those who participate in sports and those who do not. The formula for the t-test involves three main components:
  • The difference between the sample means \(\bar{x}_1 - \bar{x}_2\).
  • The variances of each sample, scaled by the sample sizes \(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\).
  • The calculated t-value, which is then compared to a critical value to determine significance.
By using this method, we aim to assess if the observed difference in manual dexterity scores is beyond what we could expect by random chance.
Grasping Type I and II Errors
When performing hypothesis testing, understanding Type I and II errors is crucial. A Type I error occurs when we wrongly reject a true null hypothesis. In other words, we claim there is an effect when there isn't one. This concept is closely tied to the significance level (alpha), which determines how probable it is to encounter such an error.
A Type II error, on the other hand, happens when we fail to reject a false null hypothesis. This leads us to overlook a real effect, concluding that no difference exists when it indeed does. The probability of making a Type II error is denoted by \(\beta\).
In our exercise, since the calculated t-value does not exceed the critical value, we risk encountering a Type II error. Understanding these errors helps us gauge the reliability and limitations of our test results.
Exploring the Power of a Test
The power of a test signifies its ability to detect an effect when there's one genuinely present. Mathematically, it is defined as \(1 - \beta\), where \(\beta\) is the probability of committing a Type II error. A test with high power is more reliable because it minimizes the chance of missing a true effect.
  • Higher power indicates a lower risk of Type II error.
  • Power depends on factors like sample size, significance level, and the true effect size.
In the context of this exercise, when the true mean difference \(\mu_1 - \mu_2\) is considered to be 3, the power of the test is approximately 0.9035. This suggests a high likelihood that the test would correctly reject the null hypothesis if the true mean difference exists.
Decoding Significance Level
The significance level, denoted by \(\alpha\), is a threshold we set to decide whether to reject the null hypothesis. Commonly set at 0.05, this level defines the probability of committing a Type I error. A lower alpha means more stringent criteria for acceptance, minimizing the risk of falsely identifying an effect.
In hypothesis testing, selecting the appropriate significance level depends on the context and potential consequences of errors. For the exercise, \(\alpha = 0.05\) indicates a 5% risk that our decision to reject the null hypothesis if incorrect could be simply due to chance.
By using significance levels, researchers balance the risk of errors with the need for reliable conclusions.

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Most popular questions from this chapter

Operators of gasoline-fueled vehicles complain about the price of gasoline in gas stations. According to the American Petroleum Institute, the federal gas tax per gallon is constant \((18.4 c\) as of January 13,2005 ), but state and local taxes vary from \(7.5 c\) to \(32.10 c\) for \(n=18\) key metropolitan areas around the country. \(^{\star}\) The total tax per gallon for gasoline at each of these 18 locations is given next. Suppose that these measurements constitute a random sample of size 18 $$\begin{aligned} &\begin{array}{llllll} 42.89 & 53.91 & 48.55 & 47.90 & 47.73 & 46.61 \end{array}\\\ &\begin{array}{llllll} 40.45 & 39.65 & 38.65 & 37.95 & 36.80 & 35.95 \end{array}\\\ &\begin{array}{llllll} 35.09 & 35.04 & 34.95 & 33.45 & 28.99 & 27.45 \end{array} \end{aligned}$$ a. Is there sufficient evidence to claim that the average per gallon gas tax is less than \(45 \%\) ? Use the \(t\) table in the appendix to bound the \(p\) -value associated with the test. b. What is the exact \(p\) -value?: c. Construct a \(95 \%\) confidence interval for the average per gallon gas tax in the United States.

High airline occupancy rates on scheduled flights are essential for profitability. Suppose that a scheduled flight must average at least \(60 \%\) occupancy to be profitable and that an examination of the occupancy rates for 12010: 00 A.M. flights from Atlanta to Dallas showed mean occupancy rate per flight of \(58 \%\) and standard deviation \(11 \% .\) Test to see if sufficient evidence exists to support a claim that the flight is unprofitable. Find the \(p\) -value associated with the test. What would you conclude if you wished to implement the test at the \(\alpha=.10\) level?

Currently, 20\% of potential customers buy soap of brand A. To increase sales, the company will conduct an extensive advertising campaign. At the end of the campaign, a sample of 400 potential customers will be interviewed to determine whether the campaign was successful. a. State \(H_{0}\) and \(H_{a}\) in terms of \(p\), the probability that a customer prefers soap brand A. b. The company decides to conclude that the advertising campaign was a success if at least 92 of the 400 customers interviewed prefer brand \(A\). Find \(\alpha\). (Use the normal approximation to the binomial distribution to evaluate the desired probability.)

The commercialism of the U.S. space program has been a topic of great interest since Dennis Tito paid \(\$ 20\) million to ride along with the Russian cosmonauts on the space shuttle. \(^{\star}\) In a survey of 500 men and 500 women, \(20 \%\) of the men and \(26 \%\) of the women responded that space should remain commercial free. a. Does statistically significant evidence exist to suggest that there is a difference in the population proportions of men and women who think that space should remain commercial free? Use a .05 level test. b. Why is a statistically significant difference in these population proportions of practical importance to advertisers?

Let \(Y_{1}, Y_{2}, \ldots, Y_{n}\) denote a random sample from a population having a Poisson distribution with mean \(\lambda_{1} .\) Let \(X_{1}, X_{2}, \ldots, X_{m}\) denote an independent random sample from a population having a Poisson distribution with mean \(\lambda_{2}\). Derive the most powerful test for testing \(H_{0}: \lambda_{1}=\lambda_{2}=2\) versus \(H_{a}: \lambda_{1}=1 / 2, \lambda_{2}=3\)

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