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James Ambercrombie believes that the average return on equity in the utility industry, µ, is greater than 10%. What are the null (H0) and alternative (Ha) hypotheses for his study?
A)
H0: µ ≤ 0.10 versus Ha: µ > 0.10.
B)
H0: µ < 0.10 versus Ha: µ > 0.10.
C)
H0: µ > 0.10 versus Ha: µ < 0.10.



This is a one-sided alternative because of the "greater than" belief.

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Which one of the following is the most appropriate set of hypotheses to use when a researcher is trying to demonstrate that a return is greater than the risk-free rate? The null hypothesis is framed as a:
A)
less than statement and the alternative hypothesis is framed as a greater than or equal to statement.
B)
less than or equal to statement and the alternative hypothesis is framed as a greater than statement.
C)
greater than statement and the alternative hypothesis is framed as a less than or equal to statement.



If a researcher is trying to show that a return is greater than the risk-free rate then this should be the alternative hypothesis. The null hypothesis would then take the form of a less than or equal to statement.

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Which one of the following best characterizes the alternative hypothesis? The alternative hypothesis is usually the:
A)
hoped-for outcome.
B)
hypothesis to be proved through statistical testing.
C)
hypothesis that is accepted after a statistical test is conducted.



The alternative hypothesis is typically the hypothesis that a researcher hopes to support after a statistical test is carried out. We can reject or fail to reject the null, not 'prove' a hypothesis.

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Jill Woodall believes that the average return on equity in the retail industry, µ, is less than 15%. What is null (H0) and alternative (Ha) hypothesis for her study?
A)
H0: µ ≥ 0.15 versus Ha: µ < 0.15.
B)
H0: µ < 0.15 versus Ha: µ = 0.15.
C)
H0: µ = 0.15 versus Ha: µ ≠ 0.15.



This is a one-sided alternative because of the “less than” belief. We expect to reject the null.

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James Ambercrombie believes that the average return on equity in the utility industry, µ, is greater than 10%. What is null (H0) and alternative (Ha) hypothesis for his study?
A)
H0: µ = 0.10 versus Ha: µ ≠ 0.10.
B)
H0: µ ≥ 0.10 versus Ha: µ < 0.10.
C)
H0: µ ≤ 0.10 versus Ha: µ > 0.10.



This is a one-sided alternative because of the “greater than” belief. We expect to reject the null.

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What is the most common formulation of null and alternative hypotheses?
A)
Less than for the null and greater than for the alternative.
B)
Equal to for the null and not equal to for the alternative.
C)
Greater than or equal to for the null and less than for the alternative.



The most common set of hypotheses will take the form of an equal to statement for the null and a not equal to statement for the alternative.

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Robert Patterson, an options trader, believes that the return on options trading is higher on Mondays than on other days. In order to test his theory, he formulates a null hypothesis. Which of the following would be an appropriate null hypothesis? Returns on Mondays are:
A)
not greater than returns on other days.
B)
greater than returns on other days.
C)
less than returns on other days.



An appropriate null hypothesis is one that the researcher wants to reject. If Patterson believes that the returns on Mondays are greater than on other days, he would like to reject the hypothesis that the opposite is true–that returns on Mondays are not greater than returns on other days.

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For a two-tailed test of hypothesis involving a z-distributed test statistic and a 5% level of significance, a calculated z-statistic of 1.5 indicates that:
A)
the null hypothesis cannot be rejected.
B)
the null hypothesis is rejected.
C)
the test is inconclusive.



For a two-tailed test at a 5% level of significance the calculated z-statistic would have to be greater than the critical z value of 1.96 for the null hypothesis to be rejected.

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A pitching machine is calibrated to deliver a fastball at a speed of 98 miles per hour. Every day, a technician samples the speed of twenty-five fastballs in order to determine if the machine needs adjustment. Today, the sample showed a mean speed of 99 miles per hour with a standard deviation of 1.75 miles per hour. Assume the population is normally distributed. At a 95% confidence level, what is the t-value in relation to the critical value?
A)
The critical value exceeds the t-value by 1.3 standard deviations.
B)
The t-value exceeds the critical value by 1.5 standard deviations.
C)
The t-value exceeds the critical value by 0.8 standard deviations.



t = (99 – 98) / (1.75 / √25) = 2.86. The critical value for a two-tailed test at the 95% confidence level with 24 degrees of freedom is ±2.06 standard deviations. Therefore, the t-value exceeds the critical value by 0.8 standard deviations.

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Ron Jacobi, manager with the Toulee Department of Natural Resources, is responsible for setting catch-and-release limits for Lake Norby, a large and popular fishing lake. For the last two months he has been sampling to determine whether the average length of Northern Pike in the lake exceeds 18 inches (using a significance level of 0.05). Assume that the p-value is 0.08. In concluding that the average size of the fish exceeds 18 inches, Jacobi:
A)
makes a Type I error.
B)
makes a Type II error.
C)
is correct.



This statement is an example of a Type I error, or rejection of a hypothesis when it is actually true (also known as the significance level of the test). Here, Ho: μ = 18 inches and Ha: μ > 18 inches. When the p-value is greater than the significance level (0.08 > 0.05), we should fail to reject the null hypothesis. Since Jacobi rejected Ho when it was true, he made a Type 1 error.
The other statements are incorrect. Type II errors occur when you fail to reject a hypothesis when it is actually false (also known as the power of the test).

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