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Chapter 9: Problem 18

The average annual total return for U.S. Diversified Equity mutual funds from 1999 to 2003 was \(4.1 \%\) (BusinessWeek, January 26,2004 ). A researcher would like to conduct a hypothesis test to see whether the returns for mid-cap growth funds over the same period are significantly different from the average for U.S. Diversified Equity funds. a. Formulate the hypotheses that can be used to determine whether the mean annual return for mid-cap growth funds differ from the mean for U.S. Diversified Equity funds. b. \(\quad\) A sample of 40 mid-cap growth funds provides a mean return of \(\bar{x}=3.4 \%\). Assume the population standard deviation for mid-cap growth funds is known from previous studies to be \(\sigma=2 \% .\) Use the sample results to compute the test statistic and \(p\) -value for the hypothesis test. c. \(\quad\) At \(\alpha=.05,\) what is your conclusion?

Short Answer

Expert verified
Reject the null hypothesis; the mean return differs significantly.

Step by step solution

01

Formulate the Hypotheses

The null hypothesis (H_0) is that the mean annual return for mid-cap growth funds is equal to the mean for U.S. Diversified Equity funds, \(\mu = 4.1\%\). The alternative hypothesis (H_a) is that the mean return for mid-cap growth funds differs from the U.S. Diversified Equity funds, \(\mu eq 4.1\%\).
02

Calculate the Test Statistic

We use the formula for the Z-test for the mean: \[ z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}} \]Given that \(\bar{x} = 3.4\%\), \(\mu = 4.1\%\), \(\sigma = 2\%\), and \(n = 40\), substitute these values: \[ z = \frac{3.4 - 4.1}{2 / \sqrt{40}} \approx -2.21 \]
03

Find the P-value

Using the Z-table (standard normal distribution table), find the probability for \(z = -2.21\). The P-value corresponds to twice the probability of \(z\) being less than -2.21 because this is a two-tailed test. The P-value \(\approx 2 \times 0.0136 = 0.0272\).
04

Make a Decision

Compare the P-value \(0.0272\) with \(\alpha = 0.05\). Since \(0.0272 < 0.05\), we reject the null hypothesis.
05

Conclusion

There is sufficient evidence to conclude that the mean annual return for mid-cap growth funds is significantly different from the mean for U.S. Diversified Equity funds.

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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 type of hypothesis test used to determine if there is a significant difference between the means of two groups. In this scenario, a researcher wants to compare the mean annual return of mid-cap growth funds to that of U.S. Diversified Equity funds. The z-test is appropriate because the population standard deviation is known and the sample size is sufficiently large (n = 40).

To conduct a z-test for means, you'll need the sample mean (\(\bar{x}\)), the population mean (\(\mu\)), the population standard deviation (\(\sigma\)), and the sample size (\(n\)). The formula for the z-test statistic is:
  • \[ z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}} \]
This formula calculates how many standard deviations away the sample mean is from the population mean.

In the example provided, the test statistic calculated is approximately \(-2.21\), which is used to assess whether the difference in means is significant.
p-value
The p-value is an essential part of hypothesis testing and helps us decide the significance of our findings. It tells us the probability of obtaining a test statistic as extreme as, or more extreme than, the one calculated, under the assumption that the null hypothesis is true.

In a two-tailed test, like the one we have, the p-value is calculated by considering the probabilities of both tails of the distribution. Using a z-table, we determine the probability for our test statistic \(z = -2.21\). This initial probability relates to one tail, so we multiply by two to account for both tails, yielding a p-value of approximately \(0.0272\).

The p-value helps us in verifying our hypothesis:
  • If the p-value is less than the significance level \(\alpha\) (0.05 in this case), it suggests the findings are statistically significant, warranting a rejection of the null hypothesis.
  • Conversely, if the p-value exceeds \(\alpha\), we fail to reject the null hypothesis.
null hypothesis
The null hypothesis (often denoted as \(H_0\)) is a default statement that there is no effect or no difference, serving as a starting point for hypothesis testing. In this study, the null hypothesis posits that the mean annual return of mid-cap growth funds is equal to that of U.S. Diversified Equity funds during the period of 1999 to 2003. Formally, it can be expressed as:
  • \[ H_0: \mu = 4.1\% \]
The null hypothesis is typically what researchers aim to test against.

A critical part of the hypothesis test is deciding whether or not to reject \(H_0\) based on the evidence provided by the sample. If the test results show that the probability of witnessing the observed data under \(H_0\) is less than a pre-determined threshold (the significance level \(\alpha\)), \(H_0\) is rejected. Otherwise, it is not rejected, suggesting that any deviation from \(H_0\) could be due to chance.

In this example, since the p-value (\(0.0272\)) is less than the significance level (\(\alpha = 0.05\)), the null hypothesis is rejected, indicating a statistically significant difference in returns.

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Most popular questions from this chapter

The Employment and Training Administration reported the U.S. mean unemployment insurance benefit of \(\$ 238\) per week (The World Almanac, 2003 ). A researcher in the state of Virginia anticipated that sample data would show evidence that the mean weekly unemployment insurance benefit in Virginia was below the national level. a. Develop appropriate hypotheses such that rejection of \(H_{0}\) will support the researcher's contention. b. For a sample of 100 individuals, the sample mean weekly unemployment insurance benefit was \(\$ 231\) with a sample standard deviation of \(\$ 80 .\) What is the \(p\) -value? c. At \(\alpha=.05,\) what is your conclusion? d. Repeat the preceding hypothesis test using the critical value approach.

Consider the following hypothesis test: \\[ \begin{array}{l} H_{0}: p=.20 \\ H_{\mathrm{a}}: p \neq .20 \end{array} \\] A sample of 400 provided a sample proportion \(\bar{p}=.175\) a. \(\quad\) Compute the value of the test statistic. b. What is the \(p\) -value? c. \(\quad\) At \(\alpha=.05,\) what is your conclusion? d. What is the rejection rule using the critical value? What is your conclusion?

The manager of an automobile dealership is considering a new bonus plan designed to increase sales volume. Currently, the mean sales volume is 14 automobiles per month. The manager wants to conduct a research study to see whether the new bonus plan increases sales volume. To collect data on the plan, a sample of sales personnel will be allowed to sell under the new bonus plan for a one-month period. a. Develop the null and alternative hypotheses most appropriate for this research situation. b. Comment on the conclusion when \(H_{0}\) cannot be rejected. c. Comment on the conclusion when \(H_{0}\) can be rejected.

Consider the following hypothesis test: \\[ \begin{array}{l} H_{0}: \mu \geq 80 \\ H_{\mathrm{a}}: \mu<80 \end{array} \\] A sample of 100 is used and the population standard deviation is \(12 .\) Compute the \(p\) -value and state your conclusion for each of the following sample results. Use \(\alpha=.01\) a. \(\quad \bar{x}=78.5\) b. \(\bar{x}=77\) c. \(\quad \bar{x}=75.5\) d. \(\bar{x}=81\)

A production line operates with a mean filling weight of 16 ounces per container. Overfilling or underfilling presents a serious problem and when detected requires the operator to shut down the production line to readjust the filling mechanism. From past data, a population standard deviation \(\sigma=.8\) ounces is assumed. A quality control inspector selects a sample of 30 items every hour and at that time makes the decision of whether to shut down the line for readjustment. The level of significance is \(\alpha=.05\) a. State the hypothesis test for this quality control application. b. If a sample mean of \(\bar{x}=16.32\) ounces were found, what is the \(p\) -value? What action would you recommend? c. If a sample mean of \(\bar{x}=15.82\) ounces were found, what is the \(p\) -value? What action would you recommend? d. Use the critical value approach. What is the rejection rule for the preceding hypothesis testing procedure? Repeat parts (b) and (c). Do you reach the same conclusion?
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