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Chapter 10: Problem 1

Independent random samples from two normal populations produced the variances listed here: Size Sample Variance Sample 16 314 a. Do the data provide sufficient evidence to indicate that \(\sigma_{1}^{2}\) differs from \(\sigma_{2}^{2}\) ? Test using \(\alpha=.05\). b. Find the approximate \(p\) -value for the test and interpret its value.

Short Answer

Expert verified
The decision regarding the null hypothesis is that we cannot reject it, as the calculated F-statistic falls within the range of the critical F-values. This means there is not enough evidence to indicate that the population variances differ significantly. The approximate p-value is greater than the significance level (0.05), which supports this finding and indicates that there is a high likelihood that the observed result could have occurred by chance alone if the population variances are equal.

Step by step solution

01

Identifying the given information

We are given the following information: Population 1: - Sample size: \(n_1 = 16\) - Sample variance: \(s_1^2 = 314\) Population 2: - Sample size: \(n_2 = 16\) - Sample variance: \(s_2^2 = 314\) Significance level: \(\alpha = 0.05\)
02

State the null and alternative hypotheses

We want to test if the variances of the two populations differ significantly. The null and alternative hypotheses can be stated as: - Null Hypothesis (\(H_0\)): The population variances are equal (\(\sigma_1^2 = \sigma_2^2\)) - Alternative Hypothesis (\(H_a\) or \(H_1\)): The population variances are not equal (\(\sigma_1^2 \neq \sigma_2^2\))
03

Perform the F-test

In order to perform the F-test, we must first calculate the F-statistic using the following formula: $$F = \frac{s_1^2}{s_2^2} = \frac{314}{314} = 1$$ Next, we need to find the critical F-values that correspond to the given significance level. Since we have a two-tailed test, we will find both the lower and upper critical values. We will use the F-distribution table and locate the values at the intersection between the numerator degrees of freedom (df1 = \(n_1 - 1\) = 15) and the denominator degrees of freedom (df2 = \(n_2 - 1\) = 15). We have: - Lower critical F-value: \(F_{0.025,15,15} = 0.393\) - Upper critical F-value: \(F_{0.975,15,15} = 2.572\)
04

Make a decision about the null hypothesis

Since the calculated F-statistic (1) falls within the range of the lower and upper critical F-values (0.393 and 2.572), we cannot reject the null hypothesis. Thus, there is not enough evidence to indicate that \(\sigma_{1}^2\) differs from \(\sigma_{2}^2\) at the 0.05 significance level.
05

Calculate the approximate p-value

To find the approximate p-value, we will use the F-distribution table or calculator. The p-value is the probability of obtaining a result as extreme or more extreme than the one observed, assuming the null hypothesis is true. Since we have a two-tailed test, we'll sum the tail probabilities on both sides of the F-distribution. However, given that our F-statistic (1) falls exactly in the middle of the distribution, the p-value is greater than the significance level of 0.05.
06

Interpret the p-value

The p-value is greater than the significance level (0.05), which means that we do not have enough evidence to reject the null hypothesis. It indicates that there is a high likelihood that our observed result is not extreme and could have occurred by chance alone if the population variances are, in fact, equal. In conclusion, the data does not provide sufficient evidence to indicate that \(\sigma_{1}^2\) differs from \(\sigma_{2}^2\), and the approximate p-value also supports this finding as it is higher than the significance level.

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Key Concepts

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

F-test
The F-test is a statistical method used to compare two population variances. In this case, we're checking if the variances of two different populations are equal. We use the formula \( F = \frac{s_1^2}{s_2^2} \), where \( s_1^2 \) and \( s_2^2 \) are the sample variances.
The F-test involves:
  • Calculating the F-statistic.
  • Comparing it to critical values from an F-distribution table.

The F-distribution is determined by degrees of freedom for both the numerator and the denominator. If the F-statistic falls outside certain critical values, it suggests that the population variances are not equal.
Significance Level
The significance level, denoted by \( \alpha \), is the probability threshold below which we reject the null hypothesis. In this problem, \( \alpha = 0.05 \), representing a 5% risk of rejecting the null hypothesis incorrectly.
Key points about significance level:
  • It's commonly set at 0.05, but this can vary depending on the study.
  • It reflects the confidence level of the test; a lower \( \alpha \) means greater confidence.

The significance level helps determine the critical values of the test, which are the cutoff points for deciding whether to reject the null hypothesis.
Null Hypothesis
In hypothesis testing, the null hypothesis \((H_0)\) represents the default or original assumption. It's a statement that there's no effect or no difference. In this exercise, the null hypothesis is that the population variances are equal: \( \sigma_1^2 = \sigma_2^2 \).
Understanding the null hypothesis:
  • It's tested against an alternative hypothesis, which suggests an effect or difference.
  • Evidence from the test is used to either reject or not reject \(H_0\).

The goal is to determine if there is significant evidence to reject the null, but not doing so doesn't prove it true, it simply implies insufficient evidence to suggest otherwise.
P-value
The p-value measures the probability of obtaining test results at least as extreme as the actual results, assuming the null hypothesis is true. In essence, it provides the smallest level of significance at which \(H_0\) can be rejected.
To interpret the p-value:
  • If it is less than \( \alpha \), reject the null hypothesis.
  • If it is greater than \( \alpha \), do not reject the null hypothesis.

In this problem, the p-value is greater than the significance level of 0.05, which means there's not enough evidence to suggest the population variances differ. A larger p-value indicates a higher likelihood that the observed result could happen by chance.

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Most popular questions from this chapter

A chemical manufacturer claims that the purity of his product never varies by more than \(2 \% .\) Five batches were tested and given purity readings of \(98.2,97.1,98.9,97.7,\) and \(97.9 \%\). a. Do the data provide sufficient evidence to contradict the manufacturer's claim? (HINT: To be generous, let a range of \(2 \%\) equal \(4 \sigma .)\) b. Find a \(90 \%\) confidence interval for \(\sigma^{2}\).

In Exercise 2.36 the number of passes completed by Brett Favre, quarterback for the Green Bay Packers, was recorded for each of the 16 regular season games in the fall of 2006 (ESPN.com): \(^{3}\) $$ \begin{array}{lllrll} 15 & 31 & 25 & 22 & 22 & 19 \\ 17 & 28 & 24 & 5 & 22 & 24 \\ 22 & 20 & 26 & 21 & & \end{array} $$ a. A stem and leaf plot of the \(n=16\) observations is shown below: Based on this plot, is it reasonable to assume that the underlying population is approximately normal, as required for the one-sample \(t\) -test? Explain. b. Calculate the mean and standard deviation for Brett Favre's per game pass completions. c. Construct a \(95 \%\) confidence interval to estimate the per game pass completions per game for Brett Favre.

Exercise 10.24 describes a dental experiment conducted to investigate the effectiveness of an oral rinse used to inhibit the growth of plaque on teeth. Subjects were divided into two groups: One group used a rinse with an anti plaque ingredient, and the control group used a rinse containing inactive ingredients. Suppose that the plaque growth on each person's teeth was measured after using the rinse after 4 hours and then again after 8 hours. If you wish to estimate the difference in plaque growth from 4 to 8 hours, should you use a confidence interval based on a paired or an unpaired analysis? Explain.

It is recognized that cigarette smoking has a deleterious effect on lung function. In a study of the effect of cigarette smoking on the carbon monoxide diffusing capacity (DL) of the lung, researchers found that current smokers had DL readings significantly lower than those of either ex smokers or nonsmokers. The carbon monoxide diffusing capacities for a random sample of \(n=20\) current smokers are listed here: $$ \begin{array}{rrrrr} 103.768 & 88.602 & 73.003 & 123.086 & 91.052 \\\ 92.295 & 61.675 & 90.677 & 84.023 & 76.014 \\ 100.615 & 88.017 & 71210 & 82.115 & 89.222 \\ 102.754 & 108.579 & 73.154 & 106.755 & 90.479 \end{array} $$ a. Do these data indicate that the mean DL reading for current smokers is significantly lower than 100 DL, the average for nonsmokers? Use \(\alpha=.01\). b. Find a \(99 \%\) upper one-sided confidence bound for the mean DL reading for current smokers. Does this bound confirm your conclusions in part a?

Maybe too good. according to tests performed by the consumer testing division of Good Housekeeping. Nutritional information provided by Kentucky Fried Chicken claims that each small bag of Potato Wedges contains 4.8 ounces of food, for a total of 280 calories. A sample of 10 orders from KFC restaurants in New York and New Jersey averaged 358 calories. \({ }^{16}\) If the standard deviation of this sample was \(s=54,\) is there sufficient evidence to indicate that the average number of calories in small bags of KFC Potato Wedges is greater than advertised? Test at the \(1 \%\) level of significance.
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