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The following data on annual rates of return were collected from 11 randomly selected stocks listed on the New York Stock Exchange ("the big board") and 12 randomly selected stocks listed on NASDAQ. Assume the population standard deviations are the same. At the .10 significance level, can we conclude that the annual rates of return are higher on the big board? $$ \begin{array}{|cc|} \hline \text { NYSE } & \text { NASDAQ } \\ \hline 15.0 & 8.8 \\ 10.7 & 6.0 \\ 20.2 & 14.4 \\ 18.6 & 19.1 \\ 19.1 & 17.6 \\ 8.7 & 17.8 \\ 17.8 & 15.9 \\ 13.8 & 17.9 \\ 22.7 & 21.6 \\ 14.0 & 6.0 \\ 26.1 & 11.9 \\ & 23.4 \\ \hline \end{array} $$

Short Answer

Expert verified
Fail to reject the null hypothesis; insufficient evidence that NYSE returns are higher.

Step by step solution

01

Calculate the means

Calculate the mean of the annual rates of return for NYSE and NASDAQ.\[ \text{Mean}_{\text{NYSE}} = \frac{15.0 + 10.7 + 20.2 + 18.6 + 19.1 + 8.7 + 17.8 + 13.8 + 22.7 + 14.0 + 26.1}{11} = 17.09 \]\[ \text{Mean}_{\text{NASDAQ}} = \frac{8.8 + 6.0 + 14.4 + 19.1 + 17.6 + 17.8 + 15.9 + 17.9 + 21.6 + 6.0 + 11.9 + 23.4}{12} = 14.92 \]
02

Calculate the standard deviations

Calculate the sample standard deviations for the NYSE and NASDAQ returns.\[ s_{\text{NYSE}} = \sqrt{\frac{\sum{(x_i - \bar{x}_{\text{NYSE}})^2}}{n - 1}} = \sqrt{\frac{((15.0-17.09)^2 + \ldots + (26.1-17.09)^2)}{10}} = 5.33 \]\[ s_{\text{NASDAQ}} = \sqrt{\frac{\sum{(x_i - \bar{x}_{\text{NASDAQ}})^2}}{n - 1}} = \sqrt{\frac{((8.8-14.92)^2 + \ldots + (23.4-14.92)^2)}{11}} = 5.62 \]
03

State the hypotheses

State the null and alternative hypotheses.* Null Hypothesis \( H_0 \): The mean annual rates of return for NYSE and NASDAQ are equal. \( \mu_{\text{NYSE}} = \mu_{\text{NASDAQ}} \)* Alternative Hypothesis \( H_a \): The mean annual rate of return for NYSE is higher than that for NASDAQ. \( \mu_{\text{NYSE}} > \mu_{\text{NASDAQ}} \)
04

Perform the t-test

Since the population standard deviations are assumed equal, perform a two-sample t-test for the difference in means.The test statistic is given by:\[ t = \frac{\bar{x}_{\text{NYSE}} - \bar{x}_{\text{NASDAQ}}}{s_p \times \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}} \]where:\[ s_p = \sqrt{\frac{(n_1-1)s_{\text{NYSE}}^2 + (n_2-1)s_{\text{NASDAQ}}^2}{n_1 + n_2 - 2}} \]Plugging in the values:\[ s_p = \sqrt{\frac{(11-1)\times 5.33^2 + (12-1)\times 5.62^2}{11 + 12 - 2}} = 5.49 \]\[ t = \frac{17.09 - 14.92}{5.49 \times \sqrt{\frac{1}{11} + \frac{1}{12}}} = 1.04 \]
05

Determine the critical value

For a significance level of 0.10 and degrees of freedom \( df = n_1 + n_2 - 2 = 21 \), use a t-distribution table to find the critical value for a one-tailed test. The critical value \( t_{0.10, 21} \) is approximately 1.323.
06

Conclusion of the test

Compare the calculated t-statistic to the critical value.Since 1.04 < 1.323, we fail to reject the null hypothesis \( H_0 \).Therefore, we do not have sufficient evidence at the 0.10 significance level to conclude that the annual rates of return on the NYSE are higher than those on NASDAQ.

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Key Concepts

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

Null Hypothesis
When conducting a t-test, one of the first steps is to formulate the null hypothesis, denoted as \( H_0 \). The null hypothesis acts as a default claim that there is no effect or no difference between groups. In the context of testing whether the annual returns of stocks listed on the NYSE are higher than those on NASDAQ, the null hypothesis states that there is no difference in the mean annual rates of return between the two stock exchanges. This can be expressed as \( \mu_{\text{NYSE}} = \mu_{\text{NASDAQ}} \).

Essentially, the null hypothesis is something we aim to test against. Our goal is to either reject it, in favor of the alternative hypothesis, or fail to reject it, which means we do not have enough statistical evidence to claim there's a difference. It's critical to remember that failing to reject the null hypothesis does not prove the null hypothesis is true—it simply means there's insufficient evidence against it.

When setting up hypotheses, the null hypothesis is usually a statement of "no effect" or "no difference," serving as a neutral ground for testing.
Alternative Hypothesis
The alternative hypothesis, denoted as \( H_a \), stands in contrast to the null hypothesis. It is a statement that indicates the presence of an effect or a difference that we expect or suspect. In our t-test scenario, the alternative hypothesis posits that the mean annual rate of return for the NYSE stocks is higher than that of the NASDAQ stocks. Mathematically, this is written as \( \mu_{\text{NYSE}} > \mu_{\text{NASDAQ}} \).

If the data provide sufficient evidence against the null hypothesis, we accept the alternative hypothesis. This means that the results support the idea that NYSE returns might indeed be higher.

The alternative hypothesis is directional since it indicates a specific direction of difference (greater than), which is crucial for determining whether to use a one-tailed or two-tailed test. In this example, a one-tailed test is appropriate as we're specifically interested in whether NYSE returns are greater.
Standard Deviation
Standard deviation is a measure of how spread out the numbers in a data set are. It provides insights into the data's variability. For our t-test, the standard deviations of both NYSE and NASDAQ data sets are crucial because they influence the calculation of the test statistic.

To compute the standard deviation of the samples, we first find the variance by taking the average of the squared differences from the mean. The formula for standard deviation is: \( s = \sqrt{\frac{\sum{(x_i - \bar{x})^2}}{n - 1}} \), where \( \bar{x} \) is the sample mean, \( n \) is the number of observations, and \( x_i \) represents each data value.

In our case, the standard deviations were calculated to be 5.33 for NYSE and 5.62 for NASDAQ. These tell us about the typical spread of returns around their respective means. Lower standard deviation indicates that the data points tend to be closer to the mean, while a higher standard deviation suggests they are more spread out. The standard deviation values feed into the pooled standard deviation \( s_p \) used in the t-test formula, affecting the t-statistic outcome.
Significance Level
The significance level, often denoted as \( \alpha \), represents the probability threshold for rejecting the null hypothesis. It's a critical value that defines how rigorous our test is—a kind of safeguard against making a Type I error, which occurs when we wrongly reject a true null hypothesis.

In this exercise, a significance level of 0.10 is chosen. This means we are willing to accept a 10% risk of declaring a difference when there actually isn't one.

During hypothesis testing, the calculated p-value is compared against the significance level:
  • If the p-value is less than the significance level, we reject the null hypothesis, indicating a significant result.
  • If the p-value is greater, we fail to reject the null hypothesis, concluding insufficient proof of significant difference.

The significance level determines the critical value for our test. In our scenario, it forms the basis for deciding whether the difference in annual returns is statistically significant or simply due to chance.

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Most popular questions from this chapter

The null and alternate hypotheses are: $$ \begin{array}{l} H_{0}: \mu_{d}=0 \\ H_{1}: \mu_{d} \neq 0 \end{array} $$ The following paired observations show the number of traffic citations given for speeding by Officer Dhondt and Officer Meredith of the South Carolina Highway Patrol for the last five months. At the .05 significance level, is there a difference in the mean number of citations given by the two officers? Note: Use the six-step hypothesis testing procedure to solve the following exercises.

Listed below are the 25 players on the opening-day roster of the 2016 New York Yankees Major League Baseball team, their salaries, and fielding positions. Sort the players into two groups, all pitchers (relief and starting) and position players (all others). Assume equal population standard deviations for the pitchers and the position players. Test the hypothesis that mean salaries of pitchers and position players are equal, using the . 01 significance level.

A computer manufacturer offers technical support that is available 24 hours a day, 7 days a week. Timely resolution of these calls is important to the company's image. For 35 calls that were related to software, technicians resolved the issues in a mean time of 18 minutes with a standard deviation of 4.2 minutes. For 45 calls related to hardware, technicians resolved the problems in a mean time of 15.5 minutes with a standard deviation of 3.9 minutes. At the .05 significance level, does it take longer to resolve software issues? What is the \(p\) -value?

Gibbs Baby Food Company wishes to compare the weight gain of infants using its brand versus its competitor's. A sample of 40 babies using the Gibbs products revealed a mean weight gain of 7.6 pounds in the first three months after birth. For the Gibbs brand, the population standard deviation of the sample is 2.3 pounds. A sample of 55 babies using the competitor's brand revealed a mean increase in weight of 8.1 pounds. The population standard deviation is 2.9 pounds. At the .05 significance level, can we conclude that babies using the Gibbs brand gained less weight? Compute the \(p\) -value and interpret it.

The president of the American Insurance Institute wants to compare the yearly costs of auto insurance offered by two leading companies. He selects a sample of 15 families, some with only a single insured driver, others with several teenage drivers, and pays each family a stipend to contact the two companies and ask for a price quote. To make the data comparable, certain features, such as the deductible amount and limits of liability, are standardized. The data for the sample of families and their two insurance quotes are reported below. At the .10 significance level, can we conclude that there is a difference in the amounts quoted? $$ \begin{array}{|lcc|} \hline \text { Family } & \begin{array}{c} \text { Midstates } \\ \text { Car Insurance } \end{array} & \begin{array}{c} \text { Gecko } \\ \text { Mutual Insurance } \end{array} \\ \hline \text { Becker } & \$ 2,090 & \$ 1,610 \\ \text { Berry } & 1,683 & 1,247 \\ \text { Cobb } & 1,402 & 2,327 \\ \text { Debuck } & 1,830 & 1,367 \\ \text { DuBrul } & 930 & 1,461 \\ \text { Eckroate } & 697 & 1,789 \\ \text { German } & 1,741 & 1,621 \\ \text { Glasson } & 1,129 & 1,914 \\ \text { King } & 1,018 & 1,956 \\ \text { Kucic } & 1,881 & 1,772 \\ \text { Meredith } & 1,571 & 1,375 \\ \text { Obeid } & 874 & 1,527 \\ \text { Price } & 1,579 & 1,767 \\ \text { Phillips } & 1,577 & 1,636 \\ \text { Tresize } & 860 & 1,188 \\ \hline \end{array} $$

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