Reading 6: hypothesis testing Flashcards
To test whether the mean of a population is greater than 20, the appropriate null hypothesis is that the population mean is:
less than 20.
greater than 20.
less than or equal to 20.
To test whether the population mean is greater than 20, the test would attempt to reject the null hypothesis that the mean is less than or equal to 20. The null hypothesis must always include the “equal to” condition. (LOS 6.a)
Which of the following statements about hypothesis testing is most accurate?
A Type II error is rejecting the null when it is actually true.
The significance level equals one minus the probability of a Type I error.
A two-tailed test with a significance level of 5% has z-critical values of ±1.96.
Rejecting the null when it is actually true is a Type I error. A Type II error is failing to reject the null hypothesis when it is false. The significance level equals the probability of a Type I error. (LOS 6.c)
For a hypothesis test with a probability of a Type II error of 60% and a probability of a Type I error of 5%, which of the following statements is most accurate?
The power of the test is 40%, and there is a 5% probability that the test statistic will exceed the critical value(s).
There is a 95% probability that the test statistic will be between the critical values if this is a two-tailed test.
There is a 5% probability that the null hypothesis will be rejected when actually true, and the probability of rejecting the null when it is false is 40%.
A Type I error is rejecting the null hypothesis when it’s true. The probability of rejecting a false null is [1 – Prob Type II] = [1 – 0.60] = 40%, which is called the power of the test. A and B are not necessarily true, since the null may be false and the probability of rejection unknown. (LOS 6.c)
If the significance level of a test is 0.05 and the probability of a Type II error is 0.15, what is the power of the test?
0.850.
0.950.
0.975.
The power of a test is 1 – P(Type II error) = 1 – 0.15 = 0.85. (LOS 6.c)
A researcher has 28 quarterly excess returns to an investment strategy and believes these returns are approximately normally distributed. The mean return on this sample is 1.645% and the standard deviation is 5.29%. For a test with a 5% significance level of the hypothesis that excess returns are less than or equal to zero, the researcher should:
reject the null hypothesis because the critical value for the test is 1.645.
not draw any conclusion because the sample size is less than 30.
fail to reject the null because the critical value is greater than 1.645.
The standard error is 5.29/sqroot28 = 1. Test statistic = 1.645/1.0 = 1.645. The critical value for t-test is greater than the critical value for a z-test at a 5% significance level (which is 1.645 for a one-tailed test), so the calculated test statistic of 1.645 must be less than the critical value for a t-test (which is 1.703 for a one-tailed test with 27 degrees of freedom) and we cannot reject the null hypothesis that mean excess return is greater than zero. (LOS 6.f)
An analyst wants to test a hypothesis concerning the population mean of monthly returns for a composite that has existed for 24 months. The analyst may appropriately use:
a t-test but not a z-test if returns for the composite are normally distributed.
either a t-test or a z-test if returns for the composite are normally distributed.
a t-test but not a z-test, regardless of the distribution of returns for the composite.
With a small sample size, a t-test may be used if the population is approximately normally distributed. If the population has a nonnormal distribution, no test statistic is available unless the sample size is large. (LOS 6.g)
From a sample of 14 observations, an analyst calculates a t-statistic to test a hypothesis that the population mean is equal to zero. If the analyst chooses a 5% significance level, the appropriate critical value is:
less than 1.80.
greater than 2.15.
between 1.80 and 2.15.
This is a two-tailed test with 14 − 1 = 13 degrees of freedom. From the t-table, 2.160 is the critical value to which the analyst should compare the calculated t-statistic. (LOS 6.g)
Which of the following assumptions is least likely required for the difference in means test based on two samples?
The two samples are independent.
The two populations are normally distributed.
The two populations have known variances.
The difference-in-means test does not require the two population variances to be known. (LOS 6.h)
William Adams wants to test whether the mean monthly returns over the last five years are the same for two stocks. If he assumes that the returns distributions are normal and have equal variances, the type of test and test statistic are best described as:
paired comparisons test, t-statistic.
paired comparisons test, F-statistic.
difference in means test, t-statistic.
Since the observations are likely dependent (both related to market returns), a paired comparisons (mean differences) test is appropriate and is based on a t-statistic. (LOS 6.h, LOS 6.i)
The appropriate test statistic for a test of the equality of variances for two normally distributed random variables, based on two independent random samples, is:
the t-test.
the F-test.
the χ2 test.
The F-test is the appropriate test. (LOS 6.j)
The appropriate test statistic to test the hypothesis that the variance of a normally distributed population is equal to 13 is:
the t-test.
the F-test.
the χ2 test.
A test of the population variance is a chi-square test. (LOS 6.j)
For a parametric test of whether a correlation coefficient is equal to zero, it is least likely that:
degrees of freedom are n – 1.
the test statistic follows a t-distribution.
the test statistic increases with a greater sample size.
Degrees of freedom are n − 2 for a test of the hypothesis that correlation is equal to zero. The test statistic increases with sample size (degrees of freedom increase) and follows a t-distribution. (LOS 6.l)
The test statistic for a Spearman rank correlation test for a sample size greater than 30 follows:
a t-distribution.
a normal distribution.
a chi-square distribution.
The test statistic for the Spearman rank correlation test follows a t-distribution. (LOS 6.l)
A contingency table can be used to test:
a null hypothesis that rank correlations are equal to zero.
whether multiple characteristics of a population are independent.
the number of p-values from multiple tests that are less than adjusted critical values.
A contingency table is used to determine whether two characteristics of a group are independent. (LOS 6.m)
