CFA Level I - Quantitative Methods- Reading 6 Hypothesis Testing
False positive and false negative results?
False positive: Type I error - true null hypothesis rejected [LOS 6.c]
F test rejection rule
1. H0: (σ1)^2=(σ2)^2 Ha: (σ1)^2≠(σ2)^2 Reject H0 if the test statistic is greater than the upper α/2 point of the F-distribution 2. H0: (σ1)^2≤(σ2)^2 Ha: (σ1)^2>(σ2)^2 Reject H0 if the test statistic is greater than the upper α point of the F-distribution [LOS 6.j]
Chi squared test rejection points
1. H0: σ^2=(σ0)^2 Ha: σ^2≠(σ0)^2 Reject H0 if the test statistic is greater than χ2(α/2) or less than χ2[1−(α/2)] 2. H0: σ2≤σ20 Ha: σ2>σ20 Reject H0 if the test statistic is greater than χ2α 3.H0: σ2≥σ20 Ha: σ2<σ20 Reject H0 if the test statistic is less than χ2(1−α) [LOS 6.j]
If the 0.01 level of significance is used, the null hypothesis that the equity risk premium is equal to 2.5% should most likely be rejected: A. only if the test statistic is above the rejection point of 2.33. B. if the test statistic is either below the rejection point of -2.33 or above the rejection point of 2.33. C. if the test statistic is either below the rejection point of -2.57 or above the rejection point of 2.58.
C. Because you are testing for equality, this is a two-sided test. As such, you should check for a 0.5% tail on the left and a 0.5% tail on the right. The proper acceptance region then is between the rejection points of -2.58 and 2.58.
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: A. paired comparisons test, t-statistic. B. paired comparisons test, F-statistic. C. difference in means test, t-statistic.
A. 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]
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 of the hypothesis that excess returns are less than or equal to zero, the researcher should: A. reject the null hypothesis because the critical value for the test is 1.645. B. not draw any conclusion because the sample size is less than 30. C. fail to reject the null because the critical value is greater than 1.645.
C. 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 1.645must 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)
A sample is taken from a normally distributed population with known variance. The observations in this sample are sorted according to an ordinal scale. To test a hypothesis regarding the sample mean, an analyst would most likely use a: A. t-test. B. z-test. C. nonparametric test.
C. When data is ranked, such as when it has been sorted according to an ordinal scale, the assumptions of parametric tests (e.g., t-test, z-test) do not hold and a nonparametric test should be used. Parametric tests require a stronger measurement scale. [LOS 6.k]
What are the critical z values for 1%, 5% and 10% significance level?
One-tailed test: 2.58, 1.96, 1.65 Two-tailed test: 2.33, 1.65, 1.28 [LOS 6.g]
What is the appropriate parametric test for the hypothesis that the population correlation equals zero, when the two variables are normally distributed?
Test statistic follows a t-distribution with n - 2 degrees of freedom. [LOS 6.l]
p-value
The smallest significance level for which the hypothesis would be rejected. [LOS 6.e]
What is the appropriate hypothesis test of the population mean when the population is normally distributed with unknown and known variance?
Unknown variance: t-test, df = n-1 (using sample standard deviation) Known variance: z-test, df = n-1 (using population standard deviation) The two tests are similar since the t-distribution approaches the z-distribution as sample size n gets large. [LOS 6.g]
An analyst takes a sample of 200 observations from a population of a stock's daily price changes. The population is normally distributed, and its variance is unknown. The sample has a mean of $1.16 and its variance is 0.8649. The degree of confidence that the population's mean price change falls between 1.1383 and 1.1817 is closest to: A 25.86% B 37.07% C 62.93%
A A confidence interval is the range within which a parameter (eg, a population mean) falls, with a certain probability. If the sample size is large (ie, at least 30) and the population variance is unknown, the z-alternative method can be used to determine a confidence interval for a population mean. Conversely, the degree of confidence, which is the probability that the sample mean falls within the confidence interval, can be estimated given a confidence interval and the sample mean. A confidence interval is the range within which a parameter (eg, a population mean) falls, with a certain probability. The square root of the sample variance is the sample standard deviation: s = (0.8649)1/2 = 0.93. Substitute known values into the formula for the z-alternative confidence interval: z = (1.1383-1.16)/(200/0.93) = -0.33 z = (1.1817-1.16)/(200/0.93) = 0.33 According to the z-table, the probability of a z-value greater than 0.33 is 1 − P(Z < 0.33) = 1 − 0.6293 = 0.3707. However, the z-table gives the one-sided probability, so the probability of being outside the interval is 2 × (0.3707) = 0.7414. Therefore, the probability of being inside the interval is 1 − 0.7414 = 25.86%. Note that a t-distribution could have applied in this question. For large sample sizes (ie, n > 30), the Central Limit Theorem allows confidence interval testing to be done using either a Z- or a t-distribution. (Choice B) 37.07% is the probability that a normal random variable is greater than 0.33. This is incorrect since the question asks for the probability that the random variable is inside both endpoints, not outside the higher endpoint. (Choice C) 62.93% is the probability that the z-value is less than 0.33. The correct answer is the probability that the z-value is between −0.33 and 0.33.
A global investment firm claims that its average annual returns exceed 20%. A consultant tests that claim by sampling 20 years of available annual returns. The sample's mean annual return is 24% with a standard deviation of 10%. If the firm's returns are normally distributed, then the most appropriate statistical decision at a 0.05 level of significance is to: A "reject" the null hypothesis. B "accept" the null hypothesis. C "not reject" the null hypothesis.
A H0: µ0 ≤ 20% Ha: µ0 > 20% The appropriate test statistic for tests population mean is t-statistics With a 20 year sample, there are n-1 = 19 degrees of freedom. Using the t-statistic table, for 19 df and probabilities of α= 0.05 we find critical value of 1.729. The computed t-value is: t = (20-24)/(10/sqrt(20)) = 1.789 The test statistic value of 1.7889 exceeds the rejection point value of 1.729, so the most appropriate statistical decision is to reject the null hypothesis (H0: µ0 ≤ 20%). This means that there is sufficient statistical evidence to support the claim that the firm's annual returns are above 20% (Ha: µ0 > 20%).
An analyst performs hypothesis tests on 20 samples from a group of small-cap stocks but is concerned about the multiple testing problem. Five samples have a p-value below the significance level of 10%, ranging from 0.0351 to 0.0784. Based on this information, the most appropriate decision is to classify these five results as: A false discoveries. B significant, since the p-values are less than 0.10. C significant, since the p-values are less than the adjusted p-values.
A The multiple testing problem is the result of a greater probability of false positives when there are many samples. To address this, the false discovery approach ranks and compares p-values with adjusted p-values, which are increasing fractions of the significance level. Only samples that pass this test are considered significant. In this scenario, five of the 20 samples have p-values ranging from 0.0351 to 0.0784, below the significance level of 0.10. The first five adjusted p-values range between 0.005 (= 0.10 × 1 / 20) and 0.025 (= 0.10 × 5 / 20). None of the p-values is smaller than the greatest adjusted p-value. Therefore, according to the false discovery approach, all five tests are false discoveries and cannot be considered significant. (Choice B) All five p-values are less than 0.10 (α), but none of them is considered significant per the stricter criteria of the false discovery approach. (Choice C) All five p-values are greater than 0.025, the greatest adjusted p-value.
An investment consultant conducts two independent random samples of five-year performance data for US and European absolute return hedge funds. Noting a return advantage of 50 bps for US managers, the consultant decides to test whether the two means are different from one another at a 0.05 level of significance. The two populations are assumed to be normally distributed with unknown but equal variances. Calculated test statistic: 0.4893 Critical value rejection points: ±1.984 The mean return for US funds is μUS, and μR is the mean return for European funds. The results of the hypothesis test most likely indicate that the: A. null hypothesis is not rejected. B. alternative hypothesis is statistically confirmed. C. difference in mean returns is statistically different from zero.
A. The calculated t-statistic value of 0.4893 falls within the bounds of the critical t-values of ±1.984. Thus, H0 cannot be rejected; the result is not statistically significant at the 0.05 level. [LOS 6.h]
In the step "stating a decision rule" in testing a hypothesis, which of the following elements most likely must be specified? A. Critical value B. Power of a test C. Value of a test statistic
A. The critical value in a decision rule is the rejection point for the test. It is the point with which the test statistic is compared to determine whether to reject the null hypothesis, which is part of the fourth step in hypothesis testing. [LOS 6.a]
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? A. 0.850. B. 0.950. C. 0.975.
A. The power of a test is 1 - P(Type II error) [LOS 6.c]
Which of the following statements is most likely correct regarding the chi-square test of independence? A. The test has a one-sided rejection region B. The null hypothesis is that the two groups are dependent C. If there are two categories, each with three levels or groups, there are six degrees of freedom
A. The test statistic comprises squared differences between the observed and expected values, so the test involves only one side, the right side. B is incorrect because the null hypothesis is that the groups are independent. C is incorrect because with three levels of groups for the two categorical variables, there are four degrees of freedom.
An asset manager is testing the correlation between stock prices of companies that provide internet streaming services and changes in the weekly jobless claims report. A sample taken from 26 weeks of data indicates a correlation of 0.40. When the hypothesis test is evaluated for correlation at both the 0.01 and 0.05 significance levels, the asset manager's most appropriate conclusion about the null hypothesis will be to: A reject it at both levels of significance. B reject it at the 0.05 level; fail to reject it at the 0.01 level. C fail to reject it at both levels of significance.
B The hypothesis test for correlation is a two-tailed t-test with (n − 2) degrees of freedom. The null hypothesis is that there is no correlation (ie, r = 0) and the alternative hypothesis is that correlation exists (ie, r ≠ 0): In this scenario, the test statistic is calculated as: t = rx[sqrt(24)/sqrt(1-r^2)] = 0.4xsqrt(24)/sqrt(84) = 2.138 The test statistic is compared to the appropriate critical value from a t-distribution table. If the absolute value of the test statistic is greater than the critical value, the null hypothesis of no correlation is rejected; otherwise, the hypothesis is not rejected. The critical value is based on n − 2, or 24 degrees of freedom. The critical value is 2.797 at a 0.01 significance level and 2.064 at a 0.05 significance level. Since this is a two-tailed test, the significance level is divided by 2 to reflect that half of the rejection area is in each tail. A common error would be to misidentify the critical values as 1.711 and 2.492 since those values appear in the 0.05 and 0.01 columns. The asset manager's most appropriate conclusion about the test at the two levels of significance will be to: Reject the null hypothesis at the 0.05 significance level since the test statistic of 2.138 is greater than the critical value of 2.064. At this level of significance, the stock prices are assumed to be correlated with weekly jobless claims (Choice C). Not reject the null hypothesis at the 0.01 significance level since the test statistic is less than the critical value of 2.797. At this level of significance, the manager cannot conclude that the stock prices and weekly jobless claims are correlated (Choice A).
An analyst wants to statistically determine whether the mean monthly returns of two large funds are significantly different. The monthly returns from the two funds are assumed to be independent of each other. The population variances are unknown but assumed to be equal. For a hypothesis test at a 0.1 level of significance, the critical value is closest to: A 1.289 B 1.658 C 1.671
B When two individual populations are normally distributed and independent, the critical value (rejection point for the hypothesis test) depends on whether the variances of the two populations are deemed equal. The critical value is found from a t-table by referencing the level of significance (0.1 given) for a two-tailed test and the degrees of freedom (df). Since the two population variances are assumed to be equal here, df is calculated as follows: df = n1 + n2 - 2= 61 + 61 -2 = 120 Since the levels of significance (α) of 0.1 correspond to a two-tailed test here (null contains an "equal to" sign), the probability in each tail equals α / 2 or 0.05. Cross-reference the t-table at 120 df and a one-tailed probability value of 0.05 (associated with α of 0.1) to obtain a critical value of 1.658.
An analyst claims that a fund has generated positive alpha. Which of the following conclusions with respect to the testing of this claim is most likely a Type II error? A. Rejecting the null hypothesis when the true alpha is greater than zero B. Accepting the null hypothesis when the true alpha is greater than zero C. Accepting the alternative hypothesis when the true alpha is less than zero
B. In this example, the "suspected" or "hoped for" condition is that the fund's true alpha is positive (greater than zero). By convention, this is established as the alternative hypothesis. The null hypothesis is that the equity risk premium is equal to or less than zero. A Type II error occurs when the null hypothesis is accepted when it should have been rejected. In this example, the alternative hypothesis should only be accepted (and the null hypothesis rejected) if the equity risk premium is greater than zero. Accepting the null hypothesis of zero or negative alpha when the fund's true alpha is positive is a Type II error. Accepting the alternative hypothesis when the true alpha is less than zero would be a Type I error. Rejecting the null hypothesis when the true alpha is greater than zero would be a correct decision. [LOS 6.c]
Which of the following statements regarding a one-tailed hypothesis test is most likely correct? A. The rejection region increases in size as the level of significance becomes smaller. B. A one-tailed test more strongly reflects the beliefs of the researcher than a two-tailed test. C. The absolute value of the rejection point is larger than that of a two-tailed test at the same level of significance.
B. One-tailed tests in which the alternative is "greater than" or "less than" represent the beliefs of the researcher more firmly than a "not equal to" alternative hypothesis.
All else equal, is specifying a smaller significance level in a hypothesis test likely to increase the probability of a Type I error and a Type II error? A. Type I error: No ; Type II error: No B. Type I error: No ; Type II error: Yes C. Type I error: Yes ; Type II error: No
B. Specifying a smaller significance level decreases the probability of a Type I error (rejecting a true null hypothesis) but increases the probability of a Type II error (not rejecting a false null hypothesis). As the level of significance decreases, the null hypothesis is less frequently rejected. [LOS 6.c]
Which of the following statements is most accurate? A Type I error: A. occurs when a false null hypothesis is not rejected. B. is less likely to be committed if the level of significance is lowered. C. is less likely to be committed if the likelihood of committing a Type II error is reduced.
B. The significance level of a test is the probability of a Type I error. If the level of significance is reduced (e.g., from 0.05 to 0.01), it is less likely that Type I errors will be committed. Type I errors occur when a true null hypothesis is rejected. Failing to reject a false null hypothesis is a description of a Type II error. Reducing the likelihood a Type I and Type II errors requires a trade-off and it is not possible to simultaneously reduce the possibility of both. If the level of significance is adjusted to reduce the likelihood of Type II errors, the likelihood of committing Type I errors will increase. [LOS 6.c]
Using historical data, a hedge fund manager designs a test of whether abnormal returns are positive on average. The test results in a p-value of 3%. The manager can most appropriately: A. reject the hypothesis that abnormal returns are less than or equal to zero, using a 1% significance level. B. reject the hypothesis that abnormal returns are less than or equal to zero, using a 5% significance level. C. conclude that the strategy produces positive abnormal returns on average, using a 5% significance level.
B. With a p-value of 3%, the manager can reject the null hypothesis (that abnormal returns are less than or equal to zero) using a significance level of 3% or higher. Although the test result is statistically significant at significance levels as small as 3%, this does not necessarily imply that the result is economically meaningful. [LOS 6.e]
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: A. less than 1.80. B. greater than 2.15. C. between 1.80 and 2.15.
B. 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)
An analyst performs a hypothesis test on the correlation of monthly returns of Fund A and Fund B. The null hypothesis is that the returns are uncorrelated with each other. Increasing the number of observations used in the test most likely results in an increase in the: A critical value. B probability of rejecting a true null hypothesis. C probability of rejecting a false null hypothesis.
C. Increasing the sample size n results in two consequences: The critical value decreases The test statistic increases The effect of both consequences is to increase the likelihood that the test statistic will exceed the critical value. This leads to a higher probability of rejecting a null hypothesis that is false (ie, making a correct decision). (Choice A) Increasing the sample size makes the critical value lower, not higher. (Choice B) As the sample size increases, the probability of rejecting a false null hypothesis increases, so the probability of rejecting a true null hypothesis (ie, a Type I error) decreases. Things to remember:The null hypothesis for a two-tailed test of correlation is that no correlation exists, and the alternative is that correlation exists. Increasing the sample size increases the possibility of making a correct decision of rejecting a false null hypothesis.
Which of the following assumptions is least likely required for the difference in means test based on two samples? A. The two samples are independent. B. The two populations are normally distributed. C. The two populations have equal variances.
C. When the variances are assumed to be unequal, we just calculate the denominator (standard error) differently and use both sample variances to calculate the t-statistic [LOS 6.h]
What is the appropriate hypothesis test to test if a population variance is equal a specific (hypothesized) value?
Chi squared test, df = n-1 Hypothesis testing of the population variance requires the use of a chi-square distributed test statistic, denoted χ2. The chi-square distribution is asymmetrical and approaches the normal distribution in shape as the degrees of freedom increase. [LOS 6.j]
What is the appropriate test for testing the independence between data based on contigency table?
Chi-square test statistic can be calculated with df = (n-1)(c-1) Ei,j, the expected number of observations in cell i,j, is: (Total in row i x total in column j)/Total for all columns and rows [LOS 6.m]
How to determine statistically significant results from multiple tests?
Compare reported p-values (ranked from lowest to highest) to the adjusted significance levels (significance level times rank / number of tests), and then count only those tests for which the p-values are less than their adjusted significance levels as rejections. [LOS 6.f]
What is the appropriate test to examine the equality/inequality of two population variances of independent samples?
F-test with df = n-1, n-2 The F-test is used under the assumption that the populations from which samples are drawn are normallterm-28y distributed and that the samples are independent. The F-test is used based on the ratio of the sample variances. The F-distribution is bounded below by 0 and defined by two values of degrees of freedom-for the numerator and for the denominator. Normally, the larger of s1 or s2 is placed in the numerator, so F is always greater than 1. [LOS 6.j]
Annie Cower is examining the earnings for two different industries. Cower suspects that the variance of earnings in the textile industry is different from the variance of earnings in the paper industry. To confirm this suspicion, Cower has looked at a sample of 31 textile manufacturers and a sample of 41 paper companies. She measured the sample standard deviation of earnings across the textile industry to be $4.30 and that of the paper industry companies to be $3.80. Cower calculates a test statistic of 1.2805. Using a 5% significance level, determine if the earnings of the textile industry have a different standard deviation than those of the paper industry.
For tests of difference between variances, the appropriate test statistic is: F=s1/s2 where s1 is the larger sample variance. Using the sample sizes for the two industries, the critical F-value for our test is found to be 1.94. This value is obtained from the table of the F-distribution for α/2=5/2=2.5% in the upper tail, with df1 = 30 and df2 =40. Since the calculated F-statistic of 1.2805 is less than the critical F-statistic of 1.94, Cower cannot reject the null hypothesis. Cower should conclude that the earnings variances of the industries are not significantly different from one another at a 5% level of significance. [LOS 6.j]
Historically, High-Return Equity Fund has advertised that its monthly returns have a standard deviation equal to 4%. This was based on estimates from the 2005-2013 period. High-Return wants to verify whether this claim still adequately describes the standard deviation of the fund's returns. High-Return collected monthly returns for the 24-month period between 2013 and 2015 and measured a standard deviation of monthly returns of 3.8%. High-Return calculates a test statistic of 20.76. Using a 5% significance level, determine if the more recent standard deviation is different from the advertised standard deviation.
H0: σ^2=4% Ha: σ^2≠4% The appropriate test statistic for tests of variance is a chi-square statistic. With a 24-month sample, there are n-1 = 23 degrees of freedom. Using the table of chi-square values, for 23 df and probabilities of α/2= 0.025 and 1-α/2=0.975 and, we find two critical values, 11.689 and 38.076. Since the computed test statistic, χ2 =20.76, falls between the two critical values, we cannot reject the null hypothesis that the variance is equal to 0.0016. The recently measured standard deviation is close enough to the advertised standard deviation that we cannot say that it is different from 4%, at a 5% level of significance. [LOS 6.j]
What is the appropriate test for comparing means of two samples which are normally distributed but dependent?
Paired comparison test: t-test with df = n-1 If the observations in the two samples both depend on some other factor, we can construct a "paired comparisons" test of whether the means of the differences between observations for the two samples are different. Dependence may result from an event that affects both sets of observations for a number of companies or because observations for two firms over time are both influenced by market returns or economic conditions. [LOS 6.i]
What is the appropriate non-parametric test for the hypothesis that the population correlation equals zero, when the two variables are normally distributed?
Spearman rank correlation coefficient: 1. Rank both X and Y from largest to smallest, assigning a rank to each value. For example, the largest observation of X has a rank value of 1, the second largest observation has a rank value of 2, and so on. 2. Calculate the difference between each pair's rank. The difference is denoted di 3. Calculate the Spearman rank correlation for a sample size of n 4. This value can be substituted into the t-test statistic calculation above. Then, we can perform a hypothesis test using n−2 degrees of freedom. [LOS 6.l]
What is the test for testing the hypothesis that the means of two normally distributed populations are equal, when the variances of the populations are unknown but assumed to be equal?
T-test with df equal to (n1 + n2 -2) If the sample means are very close together, the numerator of the t-statistic (and the t-statistic itself) are small, and we do not reject equality. If the sample means are far apart, the numerator of the t-statistic (and the t- statistic itself) are large, and we reject equality. This test is only valid for two populations that are independent and normally distributed. [LOS 6.h]
A researcher computes the sample correlation coefficient for two normally distributed random variables as 0.35, based on a sample size of 42. Determine whether to reject the hypothesis that the population correlation coefficient is equal to zero at a 5% significance level.
Test statistic = 0.35(42-2)/(sqrt(1-0.35^2)) = 2.363 Using the t-table with 42 - 2 = 40 degrees of freedom for a two-tailed test and a significance level of 5%, we can find the critical value of 2.021. Because our computed test statistic of 2.363 is greater than 2.021, we reject the hypothesis that the population mean is zero and conclude that it is not equal to zero. That is, the two populations are correlated, in this case positively. [LOS 6.l]
You have analyzed the fund performance for a 24-month period. During this period, the sample standard deviation of monthly returns is 4.2%. The fund manager claims the standard deviation of monthly returns for this strategy is less than 5.0%. Assuming the returns are normally distributed, determine if the fund manager's claim is accurate using the 0.05 level of significance.
The following hypotheses should be used. H0:σ2≥25 Ha:σ2<25 The test statistic is chi-square with 24 - 1 = 23 degrees of freedom. Reject the null hypothesis if the test statistic is less than χ2(1−0.05). The value of 13.091 can be found in a χ2-table with 23 degrees of freedom and an area of 0.95 in the right tail. χ2=[(24−1)(4.2^2)]/25=16.23 Since 16.23>13.091, you do not reject the null hypothesis. There is not enough evidence to say the fund's standard deviation is less than 5.0%. [LOS 6.j]
A researcher has gathered data on the daily returns on a portfolio of call options over a recent 250-day period. The mean daily return has been 0.1%, and the sample standard deviation of daily portfolio returns is 0.25%. The researcher believes that the mean daily portfolio return is not equal to zero. Should the hypothesis be test using t-test or z-test?
The population variance for our sample of returns is unknown. Hence, the t-distribution is appropriate. With 250 observations, however, the sample is considered to be large, so the z- distribution would also be acceptable. Because our sample is so large, the critical values for the t and z are almost identical. Hence, there is almost no difference in the likelihood of rejecting a true null. [LOS 6.g]
When your company's gizmo machine is working properly, the mean length of gizmos is 2.5 inches. However, from time to time the machine gets out of alignment and produces gizmos that are either too long or too short. When this happens, production is stopped and the machine is adjusted. To check the machine, the quality control department takes a gizmo sample each day. Today, a random sample of 49 gizmos showed a mean length of 2.49 inches. The population standard deviation is known to be 0.021 inches. The z-statistics is calculated to be -3.33. Using a 5% significance level, determine if the machine should be shut down and adjusted.
The test is two-tailed with two rejection regions, one in each tail of the standard normal distribution curve. Because the total area of both rejection regions combined is 0.05 (the significance level), the area of the rejection region in each tail is 0.025. The critical z-values for ±z0.025 are ±1.96. This means that the null hypothesis should not be rejected if the computed z-statistic lies between -1.96 and +1.96. Since z-value = -3.33, we can reject the null hypothesis. [LOS 6.g]