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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. College Room and Board Costs Room and board fees for a random sample of independent religious colleges are shown. $$\begin{array}{lllll}{7460} & {7959} & {7650} & {8120} & {7220} \\ {8768} & {7650} & {8400} & {7860} & {6782} \\ {8754} & {7443} & {9500} & {9100}\end{array}$$ Estimate the standard deviation in costs based on \(s \approx R / 4 .\) Is there sufficient evidence to conclude that the sample standard deviation differs from this estimated amount? Use \(\alpha=0.05 .\)

Short Answer

Expert verified
Calculate and compare the chi-square statistic with critical values to test variance.

Step by step solution

01

Calculate the Range

The range \(R\) is the difference between the maximum and minimum room and board costs: \[ R = \text{max}(9500, 9100, 8768, ...) - \text{min}(6782, 7220, 7443, ...). \] After calculating, \( R = 9500 - 6782 = 2718. \)
02

Estimate the Sample Standard Deviation

Use the approximation formula \( s \approx \frac{R}{4}\) to estimate the sample standard deviation. Plug in the range: \[ s \approx \frac{2718}{4} = 679.5. \]
03

State the Hypotheses

Formulate the null and alternative hypotheses. The null hypothesis \( H_0 \) is that the sample standard deviation \( s \) equals the estimated amount \( 679.5 \). The alternative hypothesis \( H_1 \) is that \( s eq 679.5 \).
04

Calculate Actual Sample Standard Deviation

Calculate the actual standard deviation of the sample data. First, find the mean: \[ \bar{x} = \frac{7460 + 7959 + ... + 9100}{14} = 8044.36. \] Then, find the sum of squared differences from the mean and divide by the sample size minus one: \[ s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}, \] where \( n = 14 \).
05

Use Chi-Square Test for Variance

To test if the sample standard deviation differs from 679.5, use the chi-square statistic: \[ \chi^2 = \frac{(n-1)s^2}{\sigma^2}, \] where \(\sigma = 679.5\). Compare the computed \(\chi^2\) value to the critical values from the chi-square distribution table for \( n-1 = 13\) degrees of freedom and \(\alpha = 0.05\).
06

Make a Decision

If the computed \(\chi^2\) falls outside the critical range, reject \( H_0 \); otherwise, do not reject \( H_0 \). Calculate \(s\) using the formula from Step 4, compare the test statistic to critical values from the chi-square distribution table, and conclude if there is evidence that \( s eq 679.5 \).

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Key Concepts

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

Chi-Square Test
The Chi-Square test is a statistical method used to determine if a sample variance significantly differs from a known or expected variance. It is particularly useful when working with normal or approximately normal distributions.

When conducting a Chi-Square test for variance, one calculates the test statistic using the formula: \[ \chi^2 = \frac{(n-1)s^2}{\sigma^2} \]where:
  • \( \chi^2 \) is the test statistic,
  • \( n \) is the sample size,
  • \( s^2 \) is the variance of the sample data,
  • \( \sigma^2 \) is the expected variance.
The degrees of freedom for this test are \( n-1 \). Once you have calculated \( \chi^2 \), you compare it to values from the Chi-Square distribution table.

By choosing a significance level (usually \( \alpha = 0.05 \)), you can find critical values in the Chi-Square table that define whether your result is statistically significant. If the calculated \( \chi^2 \) is greater or less than these critical values, you reject the null hypothesis. This indicates that the sample variance is significantly different from your expected variance.
Sample Standard Deviation
Sample standard deviation is a measure of the amount of variation or dispersion in a set of values. Unlike the population standard deviation, which measures the dispersion of all data points within a population, the sample standard deviation uses a sample subset drawn from a larger population.

To calculate the sample standard deviation, follow these steps:1. Compute the sample mean \( \bar{x} \).2. Subtract the mean from each data point to get the deviation from the mean.3. Square each of these deviations.4. Sum up all the squared deviations.5. Divide by the sample size minus one \( n-1 \) to get the variance.6. Take the square root of the variance to get the standard deviation \( s \).The formula for sample standard deviation is:\[ s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}} \]Here,
  • \( x_i \) represents each data point,
  • \( \bar{x} \) is the sample mean, and
  • \( n \) is the number of data points in the sample.
This value provides an estimate of how much individual data points deviate from the sample mean, which is crucial for hypothesis testing and confidence interval calculations.
Null and Alternative Hypotheses
Hypothesis testing is a fundamental aspect of inferential statistics. The process involves two types of hypotheses: the null hypothesis \( H_0 \) and the alternative hypothesis \( H_1 \). These represent competing claims about a population parameter.

  • Null Hypothesis \( H_0 \): This is a statement that there is no effect or no difference. In our context, it hypothesizes that the sample standard deviation equals the estimated standard deviation (e.g., \( s = 679.5 \)).
  • Alternative Hypothesis \( H_1 \): This statement contradicts \( H_0 \). It suggests there is an effect or a difference. Here, it claims that the sample standard deviation is not equal to 679.5 (\( s eq 679.5 \)).
In hypothesis testing, you assume \( H_0 \) is true unless evidence suggests otherwise. The decision to reject or fail to reject \( H_0 \) is based on a significance level \( \alpha \), usually 0.05. If the test statistic falls within the critical region defined by \( \alpha \), you reject \( H_0 \) in favor of \( H_1 \).

This approach helps in making data-driven decisions about hypotheses concerning population parameters based on sample data.

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Most popular questions from this chapter

For Exercises 5 through \(20,\) perform each of the following steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Runaways A corrections officer read that \(58 \%\) of runaways are female. He believes that the percentage is higher than \(58 .\) He selected a random sample of 90 runaways and found that 63 were female. At \(\alpha=0.05,\) can you conclude that his belief is correct?

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 \(20,\) perform each of the following steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Natural Gas Heat The Energy Information Administration reported that \(51.7 \%\) of homes in the United States were heated by natural gas. A random sample of 200 homes found that 115 were heated by natural gas. Does the evidence support the claim, or has the percent- age changed? Use \(\alpha=0.05\) and the \(P\) -value method. What could be different if the sample were taken in a different geographic area?

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?

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