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Chapter 11: Problem 18

For a projected one-tailed test, lower tail critical, at the .05 level of significance, construct two rough graphs. Each graph should show the sector in the true sampling distribution that produces a type II error and the sector that produces a correct decision. One graph should reflect the case when \(H_{0}\) really is false because the true population mean is slightly less than the hypothesized population mean, and the other graph should reflect the case when \(H_{0}\) really is false because the true population mean is appreciably less than the hypothesized population mean. (Hint: First, identify the decision rule for the hypothesized population mean, and then draw the true sampling distribution for each case.)

Short Answer

Expert verified
Two graphs are constructed to visualize the sectors in the sampling distribution that produce Type II errors and correct decisions for a one-tail test at the 0.05 level of significance. In both cases, the true population mean is less than the hypothesized mean, but to varying degrees. The critical value (decision rule) for rejecting the null hypothesis is -1.645. The areas representing correct decisions and Type II errors differ based on how much the true mean deviates from the hypothesized mean.

Step by step solution

01

Understand the Hypotheses

Start by understanding that the null hypothesis, \(H_{0}\), is the status quo, typically indicating that no effect or change is present. On the contrary, the alternative hypothesis indicates some degree of effect or change. In this exercise, we're considering a one-tailed test where we're only interested in whether the true population mean is less than the hypothesized mean, not if it's higher or equal.
02

Identify Decision Rule

The decision rule for this exercise would be: Reject \(H_{0}\) if the sample mean is less than the critical value. The critical value for a one-tail test at the 0.05 significance level is about -1.645 in the z-distribution. If the test statistic falls in the critical region (is smaller than -1.645), we reject the null and accept the alternate hypothesis that the true mean is less than hypothesized.
03

Create Graphs

Two graphs need to be created:1) True Mean is Slightly Less than Hypothesized Mean: Draw the sampling distribution with its centre slightly to the left of the hypothesized mean. Mark the critical value (-1.645). The area to the left of the critical value represents the correct decision to reject \(H_{0}\) (denoted in one colour). The area to the right of the critical value, but still under the true mean distribution, represents Type II errors, where \(H_{0}\) is not rejected when it should be (denoted in a different colour).2) True Mean is Significantly Less than Hypothesized Mean: Follow the same process as the first graph, but draw the centre of the true mean sampling distribution substantially to the left of the hypothesized mean. Note that the area of Type II error (not rejecting \(H_{0}\) when it should be rejected) will be smaller compared to the first graph, as the true mean is further away from the hypothesized mean.

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Most popular questions from this chapter

Using the .05 level of significance, an investigator retains \(H_{0}\). There is, he concludes, a probability of .95 that \(H_{0}\) is true. Comments?

Ordinarily, the investigator might not be too concerned about the low detection rate of .33 for the relatively small three-point effect of vitamin \(\mathrm{C}\) on IQ. Under special circumstances, however, this low detection rate might be unacceptable. For example, previous experimentation might have established that vitamin \(\mathrm{C}\) has many positive effects, including the reduction in the duration and severity of common colds, and no apparent negative side effects. \(^{*}\) Furthermore, huge quantities of vitamin \(\mathrm{C}\) might be available at no cost to the school district. The establishment of one more positive effect, even a fairly mild one such as a small increase in the population mean IQ, might clinch the case for supplying vitamin \(\mathrm{C}\) to all students in the district. The investigator, therefore, might wish to use a test procedure for which, if \(H_{0}\) really is false because of a small effect, the detection rate is appreciably higher than .33 . To increase the probability of detecting a false \(H_{0}\), increase the sample size. Assuming that vitamin \(\mathrm{C}\) still has only a small three-point effect on IQ, we can check the properties of the projected one-tailed test when the sample size is increased from 36 to 100 students. Recall the formula for the standard error of the mean, \(\sigma_{\bar{x}}\) namely, $$ \sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}} $$ For the original experiment with its sample size of 36 , $$ \sigma_{\bar{x}}=\frac{15}{\sqrt{36}}=\frac{15}{6}=2.5 $$ whereas for the new experiment with its sample size of 100 , $$ \sigma_{\bar{x}}=\frac{15}{\sqrt{100}}=\frac{15}{10}=1.5 $$ Clearly, any increase in sample size causes a reduction in the standard error of the mean.

For each of the following situations, indicate whether \(H_{0}\) should be retained or rejected. Given a one-tailed test, lower tail critical with \(\alpha=.01,\) and (a) \(z=-2.34\) (b) \(z=-5.13\) (c) \(z=4.04\) Given a one-tailed test, upper tail critical with \(\alpha=.05,\) and (d) \(z=2.00\) (e) \(z=-1.80\) (f) \(z=1.61\)

Give two reasons why the research hypothesis is not tested directly.

The projected hypothesis test does not fare nearly as well if \(H_{0}\) really is false because vitamin C increases the population mean IQ by only a few points - for example, by only three points. Once again, as indicated in Figure \(11.5,\) there are two different distributions of sample means: the hypothesized sampling distribution centered about the hypothesized population mean of 100 and the true sampling distribution centered about the true population mean of 103 (which reflects the three-point effect, that is, \(100+3=103\) ). After the decision rule has been constructed with the aid of the hypothesized sampling distribution, attention shifts to the true sampling distribution from which the one randomly selected sample mean actually will originate. (a) The value of the true population mean (103) dictates the location of the true sampling distribution. (b) The critical value of \(z(1.65)\) is based on the true sampling distribution. (c) Since the hypothesized population mean of 100 really is false, it would be impossible to observe a sample mean value less than or equal to 100 . (d) A correct decision would be made if the one observed sample mean has a value of \(105 .\)
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