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A financial analyst wants to compare the turnover rates, in percent, for shares of oil-related stocks versus other stocks, such as GE and IBM. She selected 32 oil-related stocks and 49 other stocks. The mean turnover rate of oil-related stocks is 31.4 percent and the population standard deviation 5.1 percent. For the other stocks, the mean rate was computed to be 34.9 percent and the population standard deviation 6.7 percent. Is there a significant difference in the turnover rates of the two types of stock? Use the .01 significance level.

Short Answer

Expert verified
Yes, there is a significant difference at the 0.01 level.

Step by step solution

01

State the Hypotheses

Define the null and alternative hypotheses for the test. The null hypothesis (H0) states that there is no difference in the turnover rates of the two types of stocks. Mathematically, it is expressed as \( H_0: \mu_1 = \mu_2 \), where \( \mu_1 \) is the mean turnover rate of oil-related stocks and \( \mu_2 \) is the mean turnover rate of other stocks. The alternative hypothesis (H1) states there is a difference, expressed as \( H_1: \mu_1 eq \mu_2 \).
02

Determine the Significance Level

The significance level for this test is given as \( \alpha = 0.01 \). This means we have a 1% chance of incorrectly rejecting the null hypothesis if it is actually true.
03

Calculate the Test Statistic

Since population standard deviations are known, use the Z-test for comparison. The formula for the Z-test is:\[Z = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}}\]where \( \bar{x}_1 = 31.4 \), \( \bar{x}_2 = 34.9 \), \( \sigma_1 = 5.1 \), \( \sigma_2 = 6.7 \), \( n_1 = 32 \), and \( n_2 = 49 \). Substituting these values gives:\[Z = \frac{31.4 - 34.9}{\sqrt{\frac{5.1^2}{32} + \frac{6.7^2}{49}}} \approx \frac{-3.5}{\sqrt{0.8125 + 0.9166}} \approx \frac{-3.5}{1.301} \approx -2.69\]
04

Find the Critical Value

For a two-tailed test with \( \alpha = 0.01 \), look up the critical values in a Z-table. The critical Z-values are approximately \( \pm 2.576 \) for a .01 significance level split into two tails.
05

Compare the Test Statistic with Critical Value

Compare the calculated Z value of -2.69 to the critical values of \( \pm 2.576 \). Since \(-2.69 < -2.576\), the test statistic falls in the rejection region of the null hypothesis.
06

Make a Decision

Since the calculated Z value is less than the critical Z value, we reject the null hypothesis. This indicates that there is a statistically significant difference between the turnover rates of oil-related stocks and other stocks at the 0.01 significance level.

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Key Concepts

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

Z-test
The Z-test is a statistical method used to determine whether there is a significant difference between the means of two populations. It is particularly useful when the population standard deviations are known and the sample size is large. In our exercise, the financial analyst wants to test if there's a significant difference between the turnover rates of oil-related and other stocks. The Z-test helps in making this decision effectively.

To apply the Z-test, one needs to know the sample means, population standard deviations, and the sample sizes. The formula used is:\[Z = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}}\]where:- \(\bar{x}_1\) and \(\bar{x}_2\) are the sample means.
- \(\sigma_1\) and \(\sigma_2\) are the population standard deviations.
- \(n_1\) and \(n_2\) are the sample sizes.

This calculation helps determine the Z-score, which can then be compared to a critical value to establish statistical significance.
Significance Level
The significance level, denoted by \(\alpha\), is a threshold set by the researcher to decide whether to reject the null hypothesis. It represents the probability of making a Type I error, which occurs when the null hypothesis is wrongly rejected. In our exercise, the analyst has chosen a significance level of 0.01 or 1%.

This low significance level indicates that the analyst is very cautious and wants to minimize the risk of incorrectly claiming a difference when none exists. At a 0.01 level, the critical Z-values for a two-tailed test are approximately \(\pm 2.576\). This means if the Z-score falls beyond these values, the null hypothesis is rejected.

Choosing a significance level depends largely on the context and the danger of making incorrect inferences. At 0.01, there's only a 1% chance of false positives, making it a rigorous test criterion for the analysis.
Population Standard Deviation
Population standard deviation is a measure of the spread or dispersion of a set of data. It gives insight into how much the individual data points differ from the population mean. In hypothesis testing, knowing the population standard deviations allows us to use the Z-test effectively.

In the example, the population standard deviations for oil-related stocks and other stocks are 5.1% and 6.7% respectively. These values are crucial in helping calculate the Z-statistic. They show the variability within each population and contribute to the standard error calculation in the Z-test formula.

Understanding these deviations is important because they provide a basis for the spread within the populations being compared. A larger standard deviation implies greater variability, which affects the confidence and reliability of the sample mean as a representation of the population mean.

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Most popular questions from this chapter

A recent study compared the time spent together by single- and dual-earner couples. According to the records kept by the wives during the study, the mean amount of time spent together watching television among the single-earner couples was 61 minutes per day, with a standard deviation of 15.5 minutes. For the dual-earner couples, the mean number of minutes spent watching television was 48.4 minutes, with a standard deviation of 18.1 minutes. At the .01 significance level, can we conclude that the single-earner couples on average spend more time watching television together? There were 15 single-earner and 12 dual-earner couples studied.

A sample of 40 observations is selected from one population with a population standard deviation of \(5 .\) The sample mean is \(102 .\) A sample of 50 observations is selected from a second population with a population standard deviation of \(6 .\) The sample mean is \(99 .\) Conduct the following test of hypothesis using the .04 significance level. $$ \begin{array}{l} H_{0}: \mu_{1}=\mu_{2} \\ H_{1}: \mu_{1} \neq \mu_{2} \end{array} $$ a. Is this a one-tailed or a two-tailed test? b. State the decision rule. c. Compute the value of the test statistic. d. What is your decision regarding \(H_{0}\) e. What is the \(p\) -value?

The Damon family owns a large grape vineyard in western New York along Lake Erie. The grapevines must be sprayed at the beginning of the growing season to protect against various insects and diseases. Two new insecticides have just been marketed: Pernod 5 and Action. To test their effectiveness, three long rows were selected and sprayed with Pernod \(5,\) and three others were sprayed with Action. When the grapes ripened, 400 of the vines treated with Pernod 5 were checked for infestation. Likewise, a sample of 400 vines sprayed with Action were checked. The results are: At the .05 significance level, can we conclude that there is a difference in the proportion of vines infested using Pernod 5 as opposed to Action?

(a) state the decision rule, (b) compute the pooled estimate of the population variance, (c) compute the test statistic, (d) state your decision about the null hypothesis, and (e) estimate the \(p\) -value. The null and alternate hypotheses are: $$ \begin{array}{l} H_{0}: \mu_{1}=\mu_{2} \\ H_{1}: \mu_{1} \neq \mu_{2} \end{array} $$ A random sample of 10 observations from one population revealed a sample mean of 23 and a sample deviation of \(4 .\) A random sample of 8 observations from another population revealed a sample mean of 26 and a sample standard deviation of \(5 .\) At the .05 significance level, is there a difference between the population means?

Mary Jo Fitzpatrick is the vice president for Nursing Services at St. Luke's Memorial Hospital. Recently she noticed in the job postings for nurses that those that are unionized seem to offer higher wages. She decided to investigate and gathered the following information. $$ \begin{array}{|lccc|} \hline & & \text { Population } & \\ \text { Group } & \text { Mean Wage } & \text { Standard Deviation } & \text { Sample Size } \\ \hline \text { Union } & \$ 20.75 & \$ 2.25 & 40 \\ \text { Nonunion } & \$ 19.80 & \$ 1.90 & 45 \\ \hline \end{array} $$ Would it be reasonable for her to conclude that union nurses earn more? Use the .02 significance level. What is the \(p\) -value?

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