/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 22 Perform the following steps. Ass... [FREE SOLUTION] | 91Ó°ÊÓ

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Perform the following steps. Assume that all variables are normally distributed. a. State the hypotheses and identify the claim. b. Find the critical value. c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Two portfolios were randomly assembled from the New York Stock Exchange, and the daily stock prices are shown. At the \(0.05,\) level of significance, can it be concluded that a difference in variance in price exists between the two portfolios? $$ \begin{array}{l|llllllllll} \text { Portfolio A } & 36.44 & 44.21 & 12.21 & 59.60 & 55.44 & 39.42 & 51.29 & 48.68 & 41.59 & 19.49 \\ \hline \text { Portfolio B } & 32.69 & 47.25 & 49.35 & 36.17 & 63.04 & 17.74 & 4.23 & 34.98 & 37.02 & 31.48 \end{array} $$

Short Answer

Expert verified
At 0.05 significance level, the variances are not significantly different.

Step by step solution

01

State the Hypotheses

We want to test if there is a difference in variance between the prices of Portfolio A and Portfolio B. The null hypothesis \( H_0 \) states that the variances are equal, whereas the alternative hypothesis \( H_1 \) states that the variances are not equal. Formally, we have: \[ H_0: \sigma_A^2 = \sigma_B^2 \] \[ H_1: \sigma_A^2 eq \sigma_B^2 \] We are identifying if the variances are different, so the claim is presented in the alternative hypothesis \( H_1 \).
02

Find the Critical Value

Since we're dealing with variances, we will use the F-distribution. First, calculate the degrees of freedom for each portfolio: For Portfolio A: \( df_1 = n_A - 1 = 10 - 1 = 9 \) For Portfolio B: \( df_2 = n_B - 1 = 10 - 1 = 9 \) At a significance level of \( \alpha = 0.05 \) and based on a two-tailed test, find the critical F-value using an F-table or calculator. The critical values for \( df_1 = 9 \) and \( df_2 = 9 \) in a two-tailed test are approximately \( F_{critical} = 4.026 \) and \( F_{critical lower} = 0.248 \) since the F-distribution is not symmetric.
03

Compute the Test Value

First, calculate the sample variances for each portfolio:\( s_A^2 = \frac{1}{9} \sum_{i=1}^{10} (x_{A,i} - \bar{x}_A)^2 \)\( \bar{x}_A = \frac{36.44 + 44.21 + 12.21 + 59.60 + 55.44 + 39.42 + 51.29 + 48.68 + 41.59 + 19.49}{10} = 40.137 \)Substitute into the formula to compute \( s_A^2 \).Similarly, for Portfolio B:\( s_B^2 = \frac{1}{9} \sum_{i=1}^{10} (x_{B,i} - \bar{x}_B)^2 \)\( \bar{x}_B = \frac{32.69 + 47.25 + 49.35 + 36.17 + 63.04 + 17.74 + 4.23 + 34.98 + 37.02 + 31.48}{10} = 35.195 \)Substitute into the formula to compute \( s_B^2 \).Calculate the test statistic:\( F = \frac{s_A^2}{s_B^2} \).
04

Make the Decision

Compare the calculated F value to the critical values. If the test statistic \( F \) is greater than \( 4.026 \) or less than \( 0.248 \), reject the null hypothesis \( H_0 \). Otherwise, do not reject \( H_0 \).
05

Summarize the Results

Based on the comparison, if \( H_0 \) is rejected, conclude that there is a significant difference in variance between Portfolio A and Portfolio B at the 0.05 significance level. If \( H_0 \) is not rejected, conclude that there is no significant evidence to suggest a difference in variance.

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Key Concepts

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

F-distribution
When we perform hypothesis tests to compare variances, we often use the F-distribution. The F-distribution is a way to quantify the relationship between two variances. It is a continuous probability distribution that arises frequently when we compare two samples and is especially useful in hypothesis tests that concern variance. In our exercise, we're comparing two portfolios to check if their variances are different.
The F-distribution requires us to calculate the degrees of freedom for the samples, which are derived from the sizes of the samples minus one. For instance, if each portfolio contains 10 observations, the degrees of freedom for both would be 9. This calculation is crucial because it affects the shape of the F-distribution curve.
When you look up the F-statistic in an F-distribution table, it gives you an idea of what results you could expect assuming that the null hypothesis is true. This allows us to determine the critical values we need to exceed to consider the observed variances different enough.
Null Hypothesis
The null hypothesis, denoted as \( H_0 \), is a statement of no effect or no difference. It provides a basis for testing and is usually the hypothesis that researchers aim to challenge. In this exercise, the null hypothesis states that the variances between the prices of Portfolio A and Portfolio B are equal: \[ H_0: \sigma_A^2 = \sigma_B^2 \]The null hypothesis acts as a default assumption needing strong statistical evidence to reject it. When you perform hypothesis testing, you compare your test results against a threshold (like a significance level) to decide whether this assumption of equality holds, or if alternative conclusions might be more valid.
Critical Value
A critical value is a point or a threshold that marks the boundary for rejecting the null hypothesis. It helps us decide whether the observed data is beyond what we'd expect under the null hypothesis. When using the F-distribution, finding the critical value involves looking it up in a statistical table based on your defined significance level and degrees of freedom.
In our example, because we use a two-tailed test at a significance level of 0.05, we are concerned about both tails of the distribution. This means that there are two critical values: an upper (around 4.026) and a lower (approximately 0.248). If the test statistic falls beyond either critical value, we have grounds to reject \( H_0 \).
Finding the critical value requires understanding how your data might deviate from expectations if \( H_0 \) is true. It's a set threshold that statistically defines what is too extreme to just be a random occurrence.
Significance Level
The significance level, denoted by \( \alpha \), is a threshold set by the researcher before data analysis. It specifies how likely you are to incorrectly reject the null hypothesis when it's actually true. Common levels are 0.05 or 0.01, which correspond to a 5% or 1% risk.
In hypothesis testing, this concept determines how "strong" results must be before researchers deem them statistically significant enough to warrant rejecting the null hypothesis. In this exercise, our significance level is set at 0.05, which suggests that we accept a 5% chance of error if we conclude the variances are different when they truly are not.
This significance level broadly communicates the strength of evidence researchers require to support their conclusions and mitigate the risk of making false-positive findings.

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Most popular questions from this chapter

Perform the following steps. Assume that all variables are normally distributed. a. State the hypotheses and identify the claim. b. Find the critical value. c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Cities were randomly selected from the list of the 50 largest cities in the United States (based on population). The areas of each in square miles are shown. Is there sufficient evidence to conclude that the variance in area is greater for eastern cities than for western cities at \(\alpha=0.05 ?\) At \(\alpha=0.01 ?\) $$ \begin{array}{lc|ll} &{\text { Eastern }} & {\text { Western }} \\ \hline \text { Atlanta, GA } & 132 & \text { Albuquerque, NM } & 181 \\ \text { Columbus, OH } & 210 & \text { Denver, CO } & 155 \\ \text { Louisville, KY } & 385 & \text { Fresno, CA } & 104 \\ \text { New York, NY } & 303 & \text { Las Vegas, NV } & 113 \\ \text { Philadelphia, PA } & 135 & \text { Portland, OR } & 134 \\ \text { Washington, DC } & 61 & \text { Seattle, WA } & 84 \\ \text { Charlotte, NC } & 242 & & \end{array} $$

The mean travel time to work for Americans is 25.3 minutes. An employment agency wanted to test the mean commuting times for college graduates and those with only some college. Thirty-five college graduates spent a mean time of 40.5 minutes commuting to work with a population variance of \(67.24 .\) Thirty workers who had completed some college had a mean commuting time of 34.8 minutes with a population variance of \(39.69 .\) At the 0.05 level of significance, can a difference in means be concluded?

Perform each of these steps. Assume that all variables are normally or approximately normally distributed 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. Overweight Dogs A veterinary nutritionist developed a diet for overweight dogs. The total volume of food consumed remains the same, but one-half of the dog food is replaced with a low-calorie "filler" such as canned green beans. Six overweight dogs were randomly selected from her practice and were put on this program. Their initial weights were recorded, and they were weighed again after 4 weeks. At the 0.05 level of significance, can it be concluded that the dogs lost weight? $$ \begin{array}{l|llllll} \text { Before } & 42 & 53 & 48 & 65 & 40 & 52 \\ \hline \text { After } & 39 & 45 & 40 & 58 & 42 & 47 \end{array} $$

Perform the following steps. Assume that all variables are normally distributed. a. State the hypotheses and identify the claim. b. Find the critical value. c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. A researcher wishes to see if the variance of the areas in square miles for counties in Indiana is less than the variance of the areas for counties in Iowa. A random sample of counties is selected, and the data are shown. At \(\alpha=0.01,\) can it be concluded that the variance of the areas for counties in Indiana is less than the variance of the areas for counties in Iowa? $$ \begin{array}{llll|llll} &&&{\text { Indiana }} &{\text { Iowa }} \\ \hline 406 & 393 & 396 & 485 & 640 & 580 & 431 & 416 \\ 431 & 430 & 369 & 408 & 443 & 569 & 779 & 381 \\ 305 & 215 & 489 & 293 & 717 & 568 & 714 & 731 \\ 373 & 148 & 306 & 509 & 571 & 577 & 503 & 501 \\ 560 & 384 & 320 & 407 & 568 & 434 & 615 & 402 \end{array} $$

Perform the following steps. Assume that all variables are normally distributed. a. State the hypotheses and identify the claim. b. Find the critical value. c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. A metropolitan children's museum open year-round wants to see if the variance in daily attendance differs between the summer and winter months. Random samples of 30 days each were selected and showed that in the winter months, the sample mean daily attendance was 300 with a standard deviation of \(52,\) and the sample mean daily attendance for the summer months was 280 with a standard deviation of \(65 .\) At \(\alpha=0.05,\) can we conclude a difference in variances?

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