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Chapter 10: Problem 69

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?

Short Answer

Expert verified
The p-value is less than 0.01, indicating a significant difference.

Step by step solution

01

Calculate the Differences

First, calculate the difference in the number of juggled balls for each subject before and after the class.\[ \begin{align*} \text{Difference for A} &= 4 - 3 = 1, \ \text{Difference for B} &= 5 - 4 = 1, \ \text{Difference for C} &= 6 - 3 = 3, \ \text{Difference for D} &= 4 - 2 = 2, \ \text{Difference for E} &= 5 - 4 = 1, \ \text{Difference for F} &= 7 - 5 = 2. \end{align*} \] This results in the differences being \([1, 1, 3, 2, 1, 2]\).
02

Calculate the Mean Difference

Calculate the mean of these differences. The formula is given by \( \bar{d} = \frac{1}{n} \sum_{i=1}^{n} d_i \), where \( d_i \) are the differences, and \( n \) is the number of subjects. \[ \bar{d} = \frac{1 + 1 + 3 + 2 + 1 + 2}{6} = \frac{10}{6} = \frac{5}{3} \approx 1.67. \]
03

Calculate the Standard Deviation of Differences

Using the formula for standard deviation of the differences \( s_d = \sqrt{\frac{\sum_{i=1}^{n} (d_i - \bar{d})^2}{n-1}} \), calculate the value.\[ \begin{align*} s_d &= \sqrt{\frac{(1-1.67)^2 + (1-1.67)^2 + (3-1.67)^2 + (2-1.67)^2 + (1-1.67)^2 + (2-1.67)^2}{5}} \ &= \sqrt{\frac{0.4489 + 0.4489 + 1.7689 + 0.1089 + 0.4489 + 0.1089}{5}} \ &= \sqrt{\frac{3.3334}{5}} \approx 0.816. \end{align*} \]
04

Conduct the t-Test

Perform a t-test for the differences with the mean \( \mu_0 = 0 \). Use the formula \( t = \frac{\bar{d} - \mu_0}{s_d/\sqrt{n}} \), where \( n = 6 \).\[ t = \frac{1.67 - 0}{0.816/\sqrt{6}} \approx \frac{1.67}{0.333} \approx 5.01. \]
05

Determine the p-value

Using a t-distribution table for 5 degrees of freedom (\( n-1 = 6-1 \)), find the p-value for \( t = 5.01 \). The p-value will be less than 0.01 since 5.01 is highly significant at the 1% level.

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Key Concepts

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

p-value
In hypothesis testing, the p-value plays a crucial role in determining the strength of the evidence against the null hypothesis. In simple terms, the p-value indicates the probability of obtaining a test result at least as extreme as the one you observed, given that the null hypothesis is true. A low p-value suggests that the observed data is unlikely under the null hypothesis, pointing towards a significant effect or change.
For the juggling study, the p-value was determined using the t-distribution, as it allows for testing mean differences when dealing with small sample sizes. Since the calculated p-value is less than the significance level of 1% (0.01), it indicates strong evidence against the null hypothesis. Therefore, we can confidently say that the juggling class made a significant impact on the participants' ability.
standard deviation
The standard deviation is a measure of the amount of variation or dispersion in a set of values. When applied to the differences in the juggling study, it tells us how spread out the differences are from the mean difference. A lower standard deviation implies that the differences are clustered closely around the mean, while a higher one suggests more widespread values.
To calculate the standard deviation of the differences in this study, we first find the mean difference, then compute how each value deviates from the mean, square these deviations, average them, and finally, take the square root. This gives us an intuitive measure of variability, which was calculated to be approximately 0.816 in this case. Understanding this concept helps in appreciating the consistency of the juggling improvements among the subjects.
t-distribution
The t-distribution is a type of probability distribution that is symmetric and bell-shaped, much like the normal distribution, but with thicker tails. This characteristic makes it more accommodating for estimating population parameters in small sample sizes, which are common in real-world experiments such as our juggling study.
In the context of this experiment, the t-distribution was used for conducting a t-test. The choice of the t-distribution is due to having a sample size of six subjects. The thick tails of the t-distribution help account for the added uncertainty inherent in small samples. Calculations are performed using a degree of freedom, which in this case, was 5 (one less than the sample size). As the sample size increases, the t-distribution approaches the standard normal distribution.
mean difference
The mean difference is the average of the differences in values between two paired groups. In the juggling study, we calculated the mean difference in the number of juggling balls before and after the training. This value provides a straightforward measure of the average improvement in juggling skills.
To determine the mean difference, we add up all the individual differences (e.g., how many more balls each subject could juggle after the class compared to before) and divide by the number of subjects. This operation provides the average effect of the juggling class, which was found to be approximately 1.67. It's a valuable metric for summarizing the overall impact of the class and serves as a central figure in performing hypothesis tests like the t-test.

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Most popular questions from this chapter

Use the following information to answer the next ten exercises. indicate which of the following choices best identifies the hypothesis test. a. independent group means, population standard deviations and/or variances known b. independent group means, population standard deviations and/or variances unknown c. matched or paired samples d. single mean e. two proportions f. single proportion A study is done to determine if students in the California state university system take longer to graduate than students enrolled in private universities. One hundred students from both the California state university system and private universities are surveyed. From years of research, it is known that the population standard deviations are 1.5811 years and one year, respectively.

Use the following information to answer the next twelve exercises. In the recent Census, three percent of the U.S. population reported being of two or more races. However, the percent varies tremendously from state to state. Suppose that two random surveys are conducted. In the first random survey, out of 1,000 North Dakotans, only nine people reported being of two or more races. In the second random survey, out of 500 Nevadans, 17 people reported being of two or more races. Conduct a hypothesis test to determine if the population percents are the same for the two states or if the percent for Nevada is statistically higher than for North Dakota. Which distribution (normal or Student's t) would you use for this hypothesis test?

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 conclusion can you draw about the juggling class?

Use the following information to answer the next ten exercises. indicate which of the following choices best identifies the hypothesis test. a. independent group means, population standard deviations and/or variances known b. independent group means, population standard deviations and/or variances unknown c. matched or paired samples d. single mean e. two proportions f. single proportion A new chocolate bar is taste-tested on consumers. Of interest is whether the proportion of children who like the new chocolate bar is greater than the proportion of adults who like it.

Use the following information to answer the next five exercises. Two metal alloys are being considered as material for ball bearings. The mean melting point of the two alloys is to be compared. 15 pieces of each metal are being tested. Both populations have normal distributions. The following table is the result. It is believed that Alloy Zeta has a different melting point. $$\begin{array}{|l|l|l|}\hline & {\text { Sample Mean Melting Temperatures }\left(^{\circ} F\right)} & {\text { Population Standard Deviation }} \\\ \hline \text { Alloy Gamma } & {800}&{95} \\ \hline \text { Alloy zeta } & {900} &{105} \\ \hline\end{array}$$ At the 1% significance level, what is your conclusion?
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