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Chapter 10: Problem 70

Use the following information to answer next five exercises. A study was conducted to test the effectiveness of a juggling class. Before the class started, six subjects juggled as many balls as they could at once. After the class, the same six subjects juggled as many balls as they could. The differences in the number of balls are calculated. The differences have a normal distribution. Test at the 1% significance level. $$ \begin{array}{|c|c|c|c|c|c|}\hline \text { Subject } & {\mathbf{A}} & {\mathbf{B}} & {\mathbf{c}} & {\mathbf{D}} & {\mathbf{E}} & {\mathbf{F}} \\\ \hline \text { Before } & {3} & {4} & {3} & {2} & {4} & {5} \\ \hline \text { After } & {4} & {5} & {6} & {4} & {5} & {7} \\ \hline\end{array} $$

Short Answer

Expert verified
The class significantly improved juggling ability at the 1% level.

Step by step solution

01

Calculate Differences

First, calculate the differences in the number of balls juggled before and after the class for each subject. Compute this as `After - Before` for each subject. - Subject A: 4 - 3 = 1 - Subject B: 5 - 4 = 1 - Subject C: 6 - 3 = 3 - Subject D: 4 - 2 = 2 - Subject E: 5 - 4 = 1 - Subject F: 7 - 5 = 2 The differences are 1, 1, 3, 2, 1, and 2.
02

Compute the Mean and Standard Deviation

Compute the mean () and standard deviation () of the differences.Mean difference (): \[\bar{x} = \frac{1 + 1 + 3 + 2 + 1 + 2}{6} = \frac{10}{6} = 1.67\]For standard deviation (), find variance first:\[s^2 = \frac{(1-1.67)^2 + (1-1.67)^2 + (3-1.67)^2 + (2-1.67)^2 + (1-1.67)^2 + (2-1.67)^2}{5}\]Calculate each squared term:- (1-1.67)^2 = 0.448- (1-1.67)^2 = 0.448- (3-1.67)^2 = 1.768- (2-1.67)^2 = 0.109- (1-1.67)^2 = 0.448- (2-1.67)^2 = 0.109Sum of squared terms = 3.33Variance (s^2) = 0.666Standard deviation (s) = \(\sqrt{0.666} \approx 0.816\)
03

Set Up the Hypotheses

Set up the null and alternative hypotheses. - Null hypothesis (H鈧�): There is no difference in juggling ability after the class (mean difference is zero). - Alternative hypothesis (H鈧�): There is an improvement in juggling ability after the class (mean difference is greater than zero).
04

Compute Test Statistic

Calculate the t-statistic for the sample of differences.\[t = \frac{\bar{x} - \mu}{s/\sqrt{n}}\]Here, \(\mu = 0\), \(\bar{x} = 1.67\), \(s = 0.816\), and \(n = 6\).\[t = \frac{1.67 - 0}{0.816/\sqrt{6}} \approx \frac{1.67}{0.333} \approx 5.01\]
05

Determine Critical Value and Compare

Find the critical t-value for a one-tailed test at  significance level with  degrees of freedom. At F: 7; 1% level and df = n - 1 = 5, from t-distribution table, critical t-value = 3.365. Compare t-statistic (5.01) with critical value (3.365). Since 5.01 > 3.365, reject the null hypothesis.

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Key Concepts

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

t-distribution
When conducting hypothesis testing in statistics, the t-distribution is a crucial concept. The t-distribution is especially useful when dealing with small sample sizes (typically less than 30). Unlike the normal distribution, which is symmetric and bell-shaped, the t-distribution has heavier tails.

This means that it is more prone to producing values that fall far from its mean. The t-distribution approaches a normal distribution as the sample size increases.

The t-distribution is often used in Student's t-tests, which assess whether there is a significant difference between the means of two groups. It helps account for the added uncertainty inherent in testing small samples. In the exercise, the t-distribution is needed to determine the critical value for the test. When sample sizes are limited, using the t-distribution instead of the normal distribution gives a more accurate measure of statistical significance. Here, with a sample size of 6, the t-distribution ensures that we are accurately assessing the effectiveness of the juggling class.
Significance Level
The significance level, denoted as \( \alpha \), is a threshold used to determine if a hypothesis test's result is statistically significant. It represents the probability of rejecting the null hypothesis when it is actually true, known as a Type I error.

Common significance levels are 1%, 5%, and 10%, depending on the desired strictness of the test. A lower significance level means a stricter criterion for testing, reducing the chance of a Type I error.

In the juggling class exercise, a 1% significance level is chosen. This low level indicates a high standard of evidence is required to reject the null hypothesis. If the calculated t-statistic exceeds the critical t-value, it suggests that the improvement in juggling skills after the class is statistically significant. This means that there is less than a 1% chance that the observed differences are due to random variation alone.
Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion of a set of values. A low standard deviation indicates that the values tend to be close to the mean, while a high standard deviation indicates that the values are spread out over a wider range.

To calculate the standard deviation, one must first determine the variance, which is the average of the squared differences from the mean. The standard deviation is then the square root of the variance.

In our juggling class context, the standard deviation of the differences provides insight into how consistent the improvements in juggling skills are across all subjects. The standard deviation calculated was approximately \( 0.816 \), suggesting that while some variability exists, the differences are generally clustered around the mean improvement of 1.67 balls. Lower variability reinforces confidence in the consistency of the class's impact.

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Most popular questions from this chapter

Use the following information to answer the next 12 exercises: The U.S. Center for Disease Control reports that the mean life expectancy was 47.6 years for whites born in 1900 and 33.0 years for nonwhites. Suppose that you randomly survey death records for people born in 1900 in a certain county. Of the 124 whites, the mean life span was 45.3 years with a standard deviation of 12.7 years. Of the 82 nonwhites, the mean life span was 34.1 years with a standard deviation of 15.6 years. Conduct a hypothesis test to see if the mean life spans in the county were the same for whites and nonwhites. Is this a test of means or proportions?

A student at a four-year college claims that mean enrollment at four鈥搚ear colleges is higher than at two鈥搚ear colleges in the United States. Two surveys are conducted. Of the 35 two鈥搚ear colleges surveyed, the mean enrollment was 5,068 with a standard deviation of 4,777. Of the 35 four-year colleges surveyed, the mean enrollment was 5,466 with a standard deviation of 8,191.

A study is done to determine if students in the California state university system take longer to graduate, on average, than students enrolled in private universities. One hundred students from both the California state university system and private universities are surveyed. Suppose that from years of research, it is known that the population standard deviations are 1.5811 years and 1 year, respectively. The following data are collected. The California state university system students took on average 4.5 years with a standard deviation of 0.8. The private university students took on average 4.1 years with a standard deviation of 0.3.

We are interested in whether children鈥檚 educational computer software costs less, on average, than children鈥檚 entertainment software. Thirty-six educational software titles were randomly picked from a catalog. The mean cost was \(31.14 with a standard deviation of \)4.69. Thirty-five entertainment software titles were randomly picked from the same catalog. The mean cost was \(33.86 with a standard deviation of \)10.87. Decide whether children鈥檚 educational software costs less, on average, than children鈥檚 entertainment software.

Use the following information to answer next five exercises. A study was conducted to test the effectiveness of a juggling class. Before the class started, six subjects juggled as many balls as they could at once. After the class, the same six subjects juggled as many balls as they could. The differences in the number of balls are calculated. The differences have a normal distribution. Test at the 1% significance level. $$ \begin{array}{|c|c|c|c|c|c|}\hline \text { Subject } & {\mathbf{A}} & {\mathbf{B}} & {\mathbf{c}} & {\mathbf{D}} & {\mathbf{E}} & {\mathbf{F}} \\\ \hline \text { Before } & {3} & {4} & {3} & {2} & {4} & {5} \\ \hline \text { After } & {4} & {5} & {6} & {4} & {5} & {7} \\ \hline\end{array} $$ What is the p-value?
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