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Chapter 8: Problem 6

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. $$ \begin{array}{lllll}{53} & {46} & {39} & {39} & {30} \\ {47} & {38} & {73} & {43} & {41}\end{array} $$

Short Answer

Expert verified
There is no sufficient evidence to conclude that the variance differs from 100.

Step by step solution

01

State the Hypotheses

We need to determine if there is sufficient evidence to show that the variance is different from 100. The null hypothesis is \(H_0: \sigma^2 = 100\), indicating that the population variance is 100. The alternative hypothesis is \(H_1: \sigma^2 eq 100\), suggesting that the variance is different from 100.
02

Calculate the Sample Variance

First, find the mean of the sample: \( \bar{x} = \frac{53 + 46 + 39 + 39 + 30 + 47 + 38 + 73 + 43 + 41}{10} = 44.9 \). Next, compute the sample variance using \( s^2 = \frac{\sum (x_i - \bar{x})^2}{n - 1} \). This gives us \( s^2 \approx 125.21 \).
03

Determine the Test Statistic

Using the chi-square test for variance, the test statistic is \( \chi^2 = \frac{(n-1)s^2}{\sigma^2} \). Substitute \( n = 10, s^2 = 125.21, \sigma^2 = 100 \) into the formula to get \( \chi^2 = \frac{9 \times 125.21}{100} \approx 11.27 \).
04

Find the Critical Value

For a two-tailed test at the significance level \( \alpha = 0.05 \) with \( n-1 = 9 \) degrees of freedom, use a chi-square distribution table to find the critical values: \( \chi^2_{0.025, 9} \approx 2.70 \) and \( \chi^2_{0.975, 9} \approx 19.02 \).
05

Compare Test Statistic to Critical Values

Check if the test statistic falls within the critical region: If \( \chi^2 < 2.70 \) or \( \chi^2 > 19.02 \), reject \( H_0 \). Since \( 11.27 \) does not fall in either critical region, do not reject \( H_0 \).
06

Conclusion

Since we do not reject the null hypothesis, there is not enough evidence to conclude that the variance of carbohydrates in fast-food entrees differs from 100 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.

Sample Variance
To understand hypothesis testing, it's essential to comprehend the concept of sample variance. When looking at a data set, like the one provided for fast-food entrees, the sample variance helps measure how much the individual carbohydrate counts deviate from the average. This tells us how spread out the data is.

For a quick guide on calculating sample variance:
  • Find the mean (average) of your sample data.
  • Subtract the mean from each data point and square the result. This measures the deviation for each point.
  • Sum all these squared deviations.
  • Divide this total by one less than the number of observations (n-1). This step is crucial since it provides an unbiased estimate of the population variance.
In this exercise, the sample variance for carbohydrate amounts was found to be approximately 125.21, after performing these calculations.
Chi-Square Test
The chi-square test is a vital statistical tool used to determine whether there is a significant difference between observed and expected data. When applied to variance, it's particularly helpful in checking if the observed variance is different from a known or expected variance.

Some key points about the chi-square test for variance:
  • It requires numerical data, like our carbohydrate counts, to be approximately normally distributed.
  • The test statistic is calculated using the formula: \[ \chi^2 = \frac{(n-1)s^2}{\sigma^2} \]where \( n \) is the sample size, \( s^2 \) is the sample variance, and \( \sigma^2 \) is the population variance hypothesized under the null hypothesis.
  • This test helps us decide whether deviations in data are due to chance or suggestive of a more intrinsic variance.
In this problem, a chi-square value of 11.27 was computed. This value helps us proceed to the next step of comparing it against critical values.
Null Hypothesis
The null hypothesis (often denoted as \(H_0\)) serves as a starting point for statistical hypothesis testing. In variance testing, it typically posits that the sample variance matches the population variance.

Key aspects to remember about the null hypothesis:
  • It's always formulated to be a statement of equality, such as \( \sigma^2 = 100 \) in our exercise, suggesting no difference exists.
  • The purpose of testing the null hypothesis is to determine if the sample data provides enough evidence to reject this default assumption.
  • If the evidence is insufficient, like in this exercise, the null hypothesis is not rejected, implying no significant difference from the expected variance.
This forms a basis for determining conclusions based on statistical evidence.
Critical Value
Critical values are pivotal in hypothesis testing as they establish the threshold beyond which the null hypothesis is rejected. These values depend on the significance level you've set for your test, often noted as \( \alpha \), and the degrees of freedom.

Considerations for understanding critical values:
  • The significance level, \( \alpha \), is typically set at 0.05, implying a 5% chance of incorrectly rejecting the null hypothesis.
  • The degrees of freedom are based on sample size, calculated as \( n - 1 \). For example, with our sample size of 10, the degrees of freedom are 9.
  • Using the chi-square distribution table and these values, we decide the boundaries, or critical regions, to determine statistically significant results.
In the provided exercise, the critical values for the two-tailed test were approximately 2.70 and 19.02. Since our test statistic did not fall outside this range, the null hypothesis was not rejected.

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Most popular questions from this chapter

In what ways is the \(t\) distribution similar to the standard normal distribution? In what ways is the \(t\) distribution different from the standard normal distribution?

What is meant by the critical region? The noncritical region?

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. Female Physicians The percentage of physicians who are women is \(27.9 \%\). In a survey of physicians employed by a large university health system, 45 of 120 randomly selected physicians were women. Is there sufficient evidence at the 0.05 level of significance to conclude that the proportion of women physicians at the university health system exceeds \(27.9 \% ?\)

For Exercises I through 25, 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 diagrams to show the critical region (or regions), and use the traditional method of hypothesis testing unless otherwise specified. Heights of 1-Year-olds The average 1 -year-old (both genders) is 29 inches tall. A random sample of 301 -year-olds in a large day care franchise resulted in the following heights. At \(\alpha=0.05,\) can it be concluded that the average height differs from 29 inches? Assume \(\sigma=2.61\) $$ \begin{array}{lllllllllll}{25} & {32} & {35} & {25} & {30} & {26.5} & {26} & {25.5} & {29.5} & {32} \\ {30} & {28.5} & {30} & {32} & {28} & {31.5} & {29} & {29.5} & {30} & {34} \\ {29} & {32} & {27} & {28} & {33} & {28} & {27} & {32} & {29} & {29.5}\end{array} $$

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. 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?
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