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Chapter 10: Problem 111

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.

Short Answer

Expert verified
Educational software does not cost less on average.

Step by step solution

01

Formulate the Hypotheses

To determine if educational software is cheaper on average, we set up the null hypothesis and alternative hypothesis. Let \( \mu_1 \) be the mean cost of educational software and \( \mu_2 \) the mean cost of entertainment software. The null hypothesis is \( H_0: \mu_1 = \mu_2 \), and the alternative hypothesis is \( H_a: \mu_1 < \mu_2 \).
02

Identify the Test Statistic

Since we are comparing the means of two independent samples with known variances, we should use the two-sample t-test for differences in means. The test statistic formula is:\[ t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \]where \( \bar{x}_1 = 31.14 \), \( s_1 = 4.69 \), \( n_1 = 36 \), \( \bar{x}_2 = 33.86 \), \( s_2 = 10.87 \), and \( n_2 = 35 \).
03

Calculate the Test Statistic

Insert the given values into the test statistic formula:\[ t = \frac{31.14 - 33.86}{\sqrt{\frac{4.69^2}{36} + \frac{10.87^2}{35}}} \]Calculate the variance parts: \( \frac{4.69^2}{36} = 0.6105 \) and \( \frac{10.87^2}{35} = 3.375 \). Sum these to get \( 3.9855 \). Compute \( \sqrt{3.9855} = 1.9954 \). Finally, compute the test statistic \( t = \frac{-2.72}{1.9954} \approx -1.363 \).
04

Determine the Critical Value and Compare

For a one-tailed test with \( \alpha = 0.05 \), degrees of freedom can be estimated using the smaller sample size minus 1, which is \( df = 34 \). The critical value of \( t \) from the t-distribution table for \( df = 34 \), \( \alpha = 0.05 \) is approximately \(-1.690 \). Since the calculated \( t \approx -1.363 \) does not exceed the critical value, we fail to reject the null hypothesis.
05

Conclusion

Since the test statistic \( t \approx -1.363 \) is not less than the critical value \(-1.690 \), we cannot conclude that educational software costs less on average than entertainment software at the 0.05 significance level.

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Key Concepts

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

Understanding Hypothesis Testing
Hypothesis testing is a method used in statistics to determine if there is enough evidence in a sample of data to support a certain belief or hypothesis about a population. In practice, this involves comparing two statements, known as hypotheses.
- **Null Hypothesis (H_0)**: This is the default position that there is no effect or difference. In the exercise, it states that educational and entertainment software costs the same on average (H_0: mu_1 = mu_2).
- **Alternative Hypothesis (H_a)**: This is what you want to prove. Here, the belief is that educational software costs less (H_a: mu_1 < mu_2).
After formulating these hypotheses, statistical tests are conducted to determine whether the null hypothesis can be rejected in favor of the alternative.
Mean Comparison in Two-Sample Tests
In statistical analysis, comparing means from two different groups is a common task, especially in assessing whether there's a significant difference in their average values. This comparison is crucial when evaluating different treatments, materials, or in our case, software costs.
- **Two-Sample T-Test**: This test specifically checks whether there is a significant difference between the means of two independent samples. Here, educational and entertainment software represent these two groups. The test accounts for the mean and variability in each sample. - **Test Statistic**: In the exercise, this is calculated using the means, standard deviations, and sample sizes of both groups.
This method is perfect for analyzing whether the software costs differ and guides in interpreting whether observed differences are meaningful statistically.
Grasping Standard Deviation
The standard deviation is a key concept when evaluating data variability and consistency within a dataset. It shows how much individual data points deviate from the mean.
- **Low Standard Deviation**: Indicates that data points tend to be very close to the mean, suggesting consistency. For educational software, this was 4.69. - **High Standard Deviation**: Suggests that data points are spread over a wider range. The entertainment software had a standard deviation of 10.87, indicating more price variability among titles.
Understanding standard deviation is essential as it affects the calculated test statistic, providing insights into the data distribution in your hypothesis test of cost difference.
Significance Level and Critical Value
The significance level in hypothesis testing defines the probability of rejecting the null hypothesis when it is true. Commonly denoted as alpha, it sets the benchmark for making statistical decisions.
- **Alpha Level (0.05)**: This is a typical choice representing a 5% risk of concluding that a difference exists when there is none. In hypothesis tests, this ensures that findings are not due to random chance.
The **critical value** is the point that the test statistic must exceed to reject the null hypothesis. It is obtained from standard statistical tables for the t-distribution. In our example, with degrees of freedom estimated based on sample sizes, a critical value of -1.690 was found for alpha = 0.05.
Evaluating the test statistic against this critical value helps in deciding whether the observed difference in software costs is statistically significant.

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Most popular questions from this chapter

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?

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.

Use the following information to answer the next 15 exercises: Indicate if the hypothesis test is for 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 It is believed that the average grade on an English essay in a particular school system for females is higher than for males. A random sample of 31 females had a mean score of 82 with a standard deviation of three, and a random sample of 25 males had a mean score of 76 with a standard deviation of four.

Use the following information for the next five exercises. Two types of phone operating system are being tested to determine if there is a difference in the proportions of system failures (crashes). Fifteen out of a random sample of 150 phones with OS1 had system failures within the first eight hours of operation. Nine out of another random sample of 150 phones with OS2 had system failures within the first eight hours of operation. OS2 is believed to be more stable (have fewer crashes) than OS1. State the null and alternative hypotheses.

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?
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