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Chapter 8: Problem 10

For Exercises 5 through \(20,\) assume that the variables are normally or approximately normally distributed. Use the traditional method of hypothesis testing unless otherwise specified. Exam Grades A statistics professor is used to having a variance in his class grades of no more than \(100 .\) He feels that his current group of students is different, and so he examines a random sample of midterm grades as shown. At \(\alpha=0.05,\) can it be concluded that the variance in grades exceeds \(100 ?\) $$ \begin{array}{lllll}{92.3} & {89.4} & {76.9} & {65.2} & {49.1} \\ {96.7} & {69.5} & {72.8} & {67.5} & {52.8} \\ {88.5} & {79.2} & {72.9} & {68.7} & {75.8}\end{array} $$

Short Answer

Expert verified
Yes, it can be concluded that the variance in grades exceeds 100 at \( \alpha = 0.05 \).

Step by step solution

01

State the Hypotheses

We are testing whether the variance of the grades exceeds 100. Hence, our null and alternative hypotheses are:- Null Hypothesis: \( H_0: \sigma^2 \leq 100 \)- Alternative Hypothesis: \( H_1: \sigma^2 > 100 \) where \( \sigma^2 \) is the population variance.
02

Collect the Data

The grades given are: 92.3, 89.4, 76.9, 65.2, 49.1, 96.7, 69.5, 72.8, 67.5, 52.8, 88.5, 79.2, 72.9, 68.7, and 75.8. This gives us a sample size of \( n = 15 \).
03

Calculate the Sample Variance

Firstly, compute the sample mean:\[ \bar{x} = \frac{\sum x_i}{n} = \frac{92.3 + 89.4 + \ldots + 75.8}{15} = 74.5 \]Then calculate the sample variance \( s^2 \):\[ s^2 = \frac{\sum (x_i - \bar{x})^2}{n - 1} \]Substitute the values to find \( s^2 \approx 186.76 \).
04

Conduct the Chi-Square Test for Variance

We use the chi-square distribution for testing the variance; the test statistic is:\[ \chi^2 = \frac{(n - 1) \cdot s^2}{\sigma_0^2}, \quad \text{where } \sigma_0^2 = 100 \]Thus:\[ \chi^2 = \frac{(15 - 1) \cdot 186.76}{100} = 26.166 \]
05

Determine the Critical Value from Chi-Square Distribution

Look up the critical value for the chi-square distribution with \( n-1 = 14 \) degrees of freedom at \( \alpha = 0.05 \) significance level. This gives a critical value of approximately 23.685.
06

Compare the Test Statistic with the Critical Value

Since the calculated \( \chi^2 = 26.166 \) is greater than the critical value \( 23.685 \), we reject the null hypothesis.

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Key Concepts

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

Chi-Square Test
A chi-square test is a statistical method used to determine if there is a significant difference between the expected and observed variance of a dataset.
In this scenario, it's used to analyze whether the variance in students' grades exceeds a given value, namely 100.The chi-square test relies on the chi-square distribution, which is helpful when dealing with categorical data, but can also be adapted for continuous data to test for variance.
  • For variance testing, the null hypothesis posits a specific value for the population variance.
  • The test statistic is calculated using the sample variance and compared against a critical value from the chi-square distribution.
The formula for computing the test statistic (\( \chi^2 \)): \[ \chi^2 = \frac{(n - 1) \cdot s^2}{\sigma_0^2} \] In this formula:
  • \( n \) is the sample size
  • \( s^2 \) is the sample variance
  • \( \sigma_0^2 \) is the hypothesized population variance
This method is particularly useful in situations where the normality condition of the dataset is approximately met, as assumed here.
Variance
Variance is a fundamental statistical measure that describes the spread or dispersion of a set of data points.It indicates how much the individual data points differ from the mean of the data. A larger variance signifies more spread out data, while a smaller variance indicates that the data points are closer to the mean.
To calculate the sample variance (\( s^2 \)), the following formula is used:\[ s^2 = \frac{\sum (x_i - \bar{x})^2}{n - 1} \]Where:
  • \( x_i \) are the data points
  • \( \bar{x} \) is the sample mean
  • \( n \) is the number of data points
Understanding variance allows us to assess the variability in the data. In hypothesis testing, we often test whether an observed variance differs significantly from a hypothesized or known value.
Null Hypothesis
The null hypothesis (\( H_0 \)) is a fundamental component in hypothesis testing. It represents the statement we aim to test and is usually a statement of 'no effect' or 'no difference'.In this exercise, the null hypothesis states that the variance of students' grades is not more than 100 (
\( H_0: \sigma^2 \leq 100 \)).The null hypothesis assumes the status quo, and any difference from this hypothesis, detected through statistical testing, needs to be likely enough (beyond the significance level, \( \alpha \)) to convince us to reject this assumption.The burden of proof is on showing that the data provide sufficient evidence to reject \( H_0 \) in favor of the alternative hypothesis, rather than proving \( H_0 \) itself.
Alternative Hypothesis
The alternative hypothesis (\( H_1 \)) is what we consider when evidence suggests that the null hypothesis (\( H_0 \)) may not be true. This hypothesis is what we typically aim to support when conducting a hypothesis test.In this particular scenario, the alternative hypothesis proposes that the variance of the students' grades is greater than 100. This is simply expressed as:\( H_1: \sigma^2 > 100 \)A key part of hypothesis testing is deciding on the significance level (\( \alpha \)), which controls the probability of a Type I error (wrongly rejecting the null hypothesis when it is true). If the test statistic shows that it's highly improbable for the sample variance to occur under \( H_0 \) conditions, then \( H_1 \) is favored, leading to rejection of \( H_0 \).The alternative hypothesis is what leads to new insights and actions based on the testing.

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Most popular questions from this chapter

For Exercises 5 through \(20,\) assume that the variables are normally or approximately normally distributed. Use the traditional method of hypothesis testing unless otherwise specified. Golf Scores A random sample of second-round golf scores from a major tournament is listed below. At \(\alpha=0.10,\) is there sufficient evidence to conclude that the population variance exceeds \(9 ?\) $$ \begin{array}{lllll}{75} & {67} & {69} & {72} & {70} \\ {66} & {74} & {69} & {74} & {71}\end{array} $$

For Exercises 5 through \(20,\) assume that the variables are normally or approximately normally distributed. Use the traditional method of hypothesis testing unless otherwise specified. Carbohydrates in Fast Foods The number of carbohydrates found in a random sample of fast-food entrees is listed. Is there sufficient evidence to conclude that the variance differs from \(100 ?\) Use the 0.05 level of significance. $$ \begin{array}{lllll}{53} & {46} & {39} & {39} & {30} \\ {47} & {38} & {73} & {43} & {41}\end{array} $$

For Exercises 7 through \(23,\) perform each of the following steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Find the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Assume that the population is approximately normally distributed. Cell Phone Bills The average monthly cell phone bill was reported to be $\$ 50.07 by the U.S. Wireless Industry. Random sampling of a large cell phone company found the following monthly cell phone charges (in dollars): $$\begin{array}{llll}{55.83} & {49.88} & {62.98} & {70.42} \\ {60.47} & {52.45} & {49.20} & {50.02} \\ {58.60} & {51.29}\end{array}$$ At the 0.05 level of significance, can it be concluded that the average phone bill has increased?

What symbols are used to represent the probabilities of type I and type II errors?

Working at Home Workers with a formal arrangement with their employer to be paid for time worked at home worked an average of 19 hours per week. A random sample of 15 mortgage brokers indicated that they worked a mean of 21.3 hours per week at home with a standard deviation of 6.5 hours. At $$\alpha=0.05,$$ is there sufficient evidence to conclude a difference? Construct a $$95 \%$$ confidence interval for the true mean number of paid working hours at home. Compare the results of your confidence interval to the conclusion of your hypothesis test and discuss the implications.
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