Search in book...
Toggle Font Controls
Create new playlist

Name your new playlist

Playlist description (optional)
Sign In

Email address

Password

Forgot Password?

or

Continue with Facebook

Continue with Google
Sign Up

Full Name

Email address

Confirm Email Address

Password

or

Continue with Facebook

Continue with Google

6.6 Large-Sample Test of Hypothesis about a Population Proportion

Inferences about population proportions (or percentages) are often made in the context of the probability p of “success” for a binomial distribution. We saw how to use largeLarge samples from binomial distributions to form confidence intervals for p in Section 7.4. We now consider tests of hypotheses about p.

Consider, for example, a method currently used by doctors to screen women for breast cancer. The method fails to detect cancer in 20% of the women who actually have the disease. Suppose a new method has been developed that researchers hope will detect cancer more accurately. This new method was used to screen a random sample of 140 women known to have breast cancer. Of these, the new method failed to detect cancer in 12 women. Does this sample provide evidence that the failure rate of the new method differs from the one currently in use?

We first view this experiment as a binomial one with 140 screened women as the trials and failure to detect breast cancer as “Success” (in binomial terminology). Let p represent the probability that the new method fails to detect the breast cancer; on the one hand, if the new method is no better than the current one, then the failure rate is $p = .2 .$ $p = .2 .$ On the other hand, if the new method is either better or worse than the current method, then the failure rate is either smaller or larger than 20%; that is, $p \neq .2 .$ $p \neq .2 .$

We can now place the problem in the context of a test of hypothesis:

\begin{array}{l} H_{0} : p = .2 (Cancer detection rate = .2.) \\ H_{a} : p \neq .2 (Cancer detection rate differs from .2 .) \end{array}

$\begin{array}{l} H_{0} : p = .2 (Cancer detection rate = .2.) \\ H_{a} : p \neq .2 (Cancer detection rate differs from .2 .) \end{array}$

Recall that theThe sample proportion $\hat{p}$ $\hat{p}$ is really just the sample mean of the outcomes of the individual binomial trials and, as such, is approximately normally distributed (for large samples) according to the Central Limit Theorem. Thus, for large samples we can use the standard normal z as the test statistic:

\begin{array}{l} T e s t s t a t i s t i c : z & = & \frac{Sample proportion - Null hypothesized proportion}{Standard deviation of sample proportion} \\ = & \frac{\hat{p} - p_{0}}{σ_{\hat{p}}} \end{array}

$\begin{array}{l} T e s t s t a t i s t i c : z & = & \frac{Sample proportion - Null hypothesized proportion}{Standard deviation of sample proportion} \\ = & \frac{\hat{p} - p_{0}}{σ_{\hat{p}}} \end{array}$

Note that we used the symbol $p_{0}$ $p_{0}$ to represent the null hypothesized value of p.

Rejection region: We use the standard normal distribution to find the appropriate rejection region for the specified value of $α .$ $α .$ Using $α = .05,$ $α = .05,$ the two-tailed rejection region is

z < - z_{α / 2} = - z_{.025} = - 1.96 or z > z_{α / 2} = z_{.025} = 1.96

$z < - z_{α / 2} = - z_{.025} = - 1.96 or z > z_{α / 2} = z_{.025} = 1.96$

(See Figure 6.19.)

Rejection region for breast cancer example

We are now prepared to calculate the value of the test statistic. Before doing so, however, we want to be sure that the sample size is large enough to ensure that the normal approximation for the sampling distribution of $\hat{p}$ $\hat{p}$ is reasonable. Recall from Section 5.4 that weWe require both np and nq to be at least 15. Because the null hypothesized value, p₀, is assumed to be the true value of p until our test procedure indicates otherwise, we require $n p_{0} \geq 15$ $n p_{0} \geq 15$ and $n q_{0} \geq 15$ $n q_{0} \geq 15$ (where $q_{0} = 1 - p_{0}$ $q_{0} = 1 - p_{0}$ ). Since $p_{0} = .2$ $p_{0} = .2$ for our test, we have

n p_{0} = (140) (.2) = 28

$n p_{0} = (140) (.2) = 28$

and

n q_{0} = (140) (.8) = 112

$n q_{0} = (140) (.8) = 112$

Therefore, the normal distribution will provide a reasonable approximation for the sampling distribution of $\hat{p} .$ $\hat{p} .$

Returning to the hypothesis test at hand, the proportion of the screenings that failed to detect breast cancer is

\hat{p} = \frac{12}{140} = .086

$\hat{p} = \frac{12}{140} = .086$

Finally, we calculate the number of standard deviations (the z value) between the sampled and hypothesized values of the binomial proportion:

z = \frac{\hat{p} - p 0}{σ_{\hat{p}}} = \frac{\hat{p} - p 0}{\sqrt{p_{0} q_{0} / n}} = \frac{.086 - .2}{\sqrt{(.2) (.8) / 140}} = \frac{- .114}{.034} = - 3.35

$z = \frac{\hat{p} - p 0}{σ_{\hat{p}}} = \frac{\hat{p} - p 0}{\sqrt{p_{0} q_{0} / n}} = \frac{.086 - .2}{\sqrt{(.2) (.8) / 140}} = \frac{- .114}{.034} = - 3.35$

The implication is that the observed sample proportion is (approximately) 3.35 standard deviations below the null-hypothesized proportion, .2 (Figure 6.19). Therefore, we reject the null hypothesis, concluding at the .05 level of significance that the true failure rate of the new method for detecting breast cancer differs from .20. Since $\hat{p} = .086,$ $\hat{p} = .086,$ it appears that the new method is better (i.e., has a smaller failure rate) than the method currently in use. (To estimate the magnitude of the failure rate for the new method, a confidence interval can be constructed.)

The test of hypothesis about a population proportion p is summarized in the next box. Note that the procedure is entirely analogous to that used for conducting large-sample tests about a population mean.

Large-Sample Test of Hypothesis about `p`: Normal (`z`) Statistic

Test statistic:	$z_{c} = \frac{(\hat{p} - p_{0})}{σ_{\hat{p}}} = \frac{(\hat{p} - p_{0})}{\sqrt{(p_{0} q_{0} / n)}}$ $z_{c} = \frac{(\hat{p} - p_{0})}{σ_{\hat{p}}} = \frac{(\hat{p} - p_{0})}{\sqrt{(p_{0} q_{0} / n)}}$
	One-Tailed Tests	Two-Tailed Test
	$H_{0} : p = p_{0}$ $H_{0} : p = p_{0}$	$H_{0} : p = p_{0}$ $H_{0} : p = p_{0}$
	$H_{a} : p > p_{0}$ $H_{a} : p > p_{0}$	$H_{a} : p \neq p_{0}$ $H_{a} : p \neq p_{0}$
Rejection region:	$z_{c} > z_{α}$ $z_{c} > z_{α}$	$\| z_{c} \| > z_{α / 2}$ $\| z_{c} \| > z_{α / 2}$
p-value:	$P (z < z_{c})$ $P (z < z_{c})$ $P (z > z_{c})$ $P (z > z_{c})$	$2 P (z > z_{c)}$ $2 P (z > z_{c)}$ if z_c is positive
		$2 P (t < z_{c)}$ $2 P (t < z_{c)}$ if z_c is negative
Decision: Reject `H`₀ if $α > p - v a l u e$ $α > p - v a l u e$ or if test statistic (z_c) falls in rejection region where $P (z > z_{α}) = α, P (z > z_{α / 2}) = α / 2$ $P (z > z_{α}) = α, P (z > z_{α / 2}) = α / 2$ , and $α = P (Type I error) = P (Reject H_{0} \| H_{0} true) .$ $α = P (Type I error) = P (Reject H_{0} \| H_{0} true) .$

[Note: The symbol for the numerical value assigned to p under the null hypothesis is p₀.]

Conditions Required for a Valid Large-Sample Hypothesis Test for `p`

A random sample is selected from a binomial population.
The sample size n is large. (This condition will be satisfied if both $n p_{0} \geq 15$ $n p_{0} \geq 15$ and $n q_{0} \geq 15$ $n q_{0} \geq 15$ .)

Example 6.10 A Hypothesis Test for p—Proportion of Defective Batteries

Problem

The reputations (and hence sales) of many businesses can be severely damaged by shipments of manufactured items that contain a large percentage of defectives. For example, a manufacturer of alkaline batteries may want to be reasonably certain that less than 5% of its batteries are defective. Suppose 300 batteries are randomly selected from a very large shipment; each is tested and 10 defective batteries are found. Does this outcome provide sufficient evidence for the manufacturer to conclude that the fraction defective in the entire shipment is less than .05? Use $α = .01 .$ $α = .01 .$

Solution

The objective of the sampling is to determine whether there is sufficient evidence to indicate that the fraction defective, p, is less than .05. Consequently, we will test the null hypothesis that $p = .05$ $p = .05$ against the alternative hypothesis that $p < .05 .$ $p < .05 .$ The elements of the test are

$\begin{array}{l} H_{0} : p = .05 (Fraction of defective batteries equals .05.) \\ H_{a} : p < .05 (Fraction of defective batteries is less than .05.) \\ T e s t s t a t i s t i c : z = \frac{\hat{p} - p_{0}}{σ_{\hat{p}}} \\ R e j e c t i o n r e g i o n : z < - z_{.01} = - 2.33 \end{array}$ $\begin{array}{l} H_{0} : p = .05 (Fraction of defective batteries equals .05.) \\ H_{a} : p < .05 (Fraction of defective batteries is less than .05.) \\ T e s t s t a t i s t i c : z = \frac{\hat{p} - p_{0}}{σ_{\hat{p}}} \\ R e j e c t i o n r e g i o n : z < - z_{.01} = - 2.33 \end{array}$

(see Figure 6.20)

Figure 6.20

Rejection region for Example 6.10

Before conducting the test, we check to determine whether the sample size is large enough to use the normal approximation to the sampling distribution $\hat{p} .$ $\hat{p} .$ Since $n p_{0} = (300) (.05) = 15$ $n p_{0} = (300) (.05) = 15$ and $n q_{0} = (300) (.95) = 285$ $n q_{0} = (300) (.95) = 285$ are both at least 15, the normal approximation will be adequate.

We now calculate the test statistic:

$z = \frac{\hat{p} - .05}{σ_{\hat{p}}} = \frac{(10 / 300) - .05}{\sqrt{p_{0} q_{0} / n}} = \frac{.03333 - .05}{\sqrt{p_{0} q_{0} / 300}}$ $z = \frac{\hat{p} - .05}{σ_{\hat{p}}} = \frac{(10 / 300) - .05}{\sqrt{p_{0} q_{0} / n}} = \frac{.03333 - .05}{\sqrt{p_{0} q_{0} / 300}}$

Notice that we use $p_{0}$ $p_{0}$ to calculate $σ_{\hat{p}}$ $σ_{\hat{p}}$ because, in contrast to calculating $σ_{\hat{p}}$ $σ_{\hat{p}}$ for a confidence interval, the test statistic is computed on the assumption that the null hypothesis is true—that is, $p = p_{0} .$ $p = p_{0} .$ Therefore, substituting the values for $\hat{p}$ $\hat{p}$ and $p_{0}$ $p_{0}$ into the z statistic, we obtain

$z \approx \frac{- .01667}{\sqrt{(.05) (.95) / 300}} = \frac{- .01667}{.0126} = - 1.32$ $z \approx \frac{- .01667}{\sqrt{(.05) (.95) / 300}} = \frac{- .01667}{.0126} = - 1.32$

As shown in Figure 6.20, the calculated z-value does not fall into the rejection region. Therefore, there is insufficient evidence at the .01 level of significance to indicate that the shipment contains less than 5% defective batteries.

Now Work Exercise 6.76a–b

Example 6.11 Finding the `p`-Value for a Test about a Population Proportion `p`

Problem

In Example 6.10, we found that we did not have sufficient evidence at the $α = .01$ $α = .01$ level of significance to indicate that the fraction defective, p, of alkaline batteries was less than $p = .05 .$ $p = .05 .$ How strong was the weight of evidence favoring the alternative hypothesis $(H_{a} : p < .05) ?$ $(H_{a} : p < .05) ?$ Find the observed significance level (p-value) for the test.

Solution

The computed value of the test statistic z was $z = - 1.32.$ $z = - 1.32.$ Therefore, for this lower-tailed test, the observed significance level is

$p - v a l u e = P (z \leq - 1.32)$ $p - v a l u e = P (z \leq - 1.32)$

This lower-tail area is shown in Figure 6.21. The area between $z = 0$ $z = 0$ and $z = 1.32$ $z = 1.32$ is given in Table II in Appendix B as .4066. Therefore, the observed significance level is $.5 - .4066 = .0934 .$ $.5 - .4066 = .0934 .$

Figure 6.21

The observed significance level for Example 6.11

Note: The p-value can also be obtained with statistical software. The MINITAB printout shown in Figure 6.22 gives the p-value (highlighted).

Figure 6.22

MINITAB lower-tailed test of p, Example 6.11

Look Back

Although we did not reject $H_{0} : p = .05$ $H_{0} : p = .05$ at $α = .01,$ $α = .01,$ the probability of observing a z-value as small as or smaller than $- 1.32$ $- 1.32$ is only .093 if in fact $H_{0}$ $H_{0}$ is true. Therefore, we would reject $H_{0}$ $H_{0}$ if we had selected $α = .10$ $α = .10$ prior to sampling (since the observed significance level is less than .10). This point reminds us of the importance of our choice of $α$ $α$ prior to collecting the sample data.

Now Work Exercise 6.76c

Small samples

Since most surveys and studies employ large samples, the large-sample testing procedure based on the normal (z) statistic presented here will be appropriate for making inferences about a population proportion. However, in the case of small samples (where either np₀ or nq₀ is less than 15), tests for a population proportion based on the z-statistic may not be valid—especially when conducting one-tailed tests. A test of proportions that can be applied to small samples utilizes the binomial, rather than the normal, distribution. These are called exact binomial tests due to the fact that the exact (rather than approximate) p-value for the test is computed based on the binomial distribution.

For example, consider testing H₀: $p = .6$ $p = .6$ against H_a: $p < .6$ $p < .6$ , where p is the true proportion of all voters in a large city that favor a certain mayoral candidate. Suppose that in a poll of $n = 20$ $n = 20$ voters (randomly selected from all registered voters in the city), $x = 10$ $x = 10$ favor the mayoral candidate. Is this sufficient evidence at $α = .05$ $α = .05$ to reject the null hypothesis?

First, note that $p_{0} = .6$ $p_{0} = .6$ and $n p_{0} = 20 (.6) = .12$ $n p_{0} = 20 (.6) = .12$ is less than 15; consequently, we require a small-sample test of binomial proportions. Now, we observed $x = 10$ $x = 10$ voters in favor of the mayoral candidate. Then, by definition (see p. 381), the observed significance level (p-value) of this lower-tailed test will be equal to the probability of observing 10 or fewer voters in favor of the candidate, given that the true population proportion in favor of the candidate is $p = .6$ $p = .6$ . That is,

p - value = P (x \leq 10 | p = .6)

$p - value = P (x \leq 10 | p = .6)$

In Example 4.13 (p. 190), we found this binomial probability using MINITAB. The MINITAB printout is reproducedfor the binomial probability is shown in Figure 6.23. You can see that $p - v a l u e = .245$ $p - v a l u e = .245$ . Since $α = .05 < p - v a l u e$ $α = .05 < p - v a l u e$ , we fail to reject the null hypothesis.

MINITAB Output with `p`-Value for Exact Binomial Test

All three statistical software packages presented in this text—MINITAB, SAS, and SPSS—have procedures for conducting exact binomial tests in the case of small samples.

Statistics in Action Revisited

Testing a Population Proportion in the KLEENEX^® Survey

In the previous “Statistics in Action Revisited” (p. 327), we investigated Kimberly-Clark Corporation’s assertion that the company should put more than 60 tissues in an anti-viral box of KLEENEX^® tissues. We did this by testing the claim that the mean number of tissues used by a person with a cold is $μ = 60,$ $μ = 60,$ using data collected from a survey of 250 KLEENEX^® users. Another approach to the problem is to consider the proportion of KLEENEX^® users who use more than 60 tissues when they have a cold. Now the population parameter of interest is p, the proportion of all KLEENEX^® users who use more than 60 tissues when they have a cold.

Kimberly-Clark Corporation’s belief that the company should put more than 60 tissues in an anti-viral box will be supported if over half of the KLEENEX^® users surveyed use more than 60 tissues (i.e., if $p > .5$ $p > .5$ ). Is there evidence to indicate that the population proportion exceeds .5? To answer this question, we set up the following null and alternative hypotheses:

H_{0} : p = .5 H_{a} : p > .5

$H_{0} : p = .5 H_{a} : p > .5$

Recall that the survey results for the 250 sampled KLEENEX^® users are stored in the TISSUES data file. In addition to the number of tissues used by each person, the file contains a qualitative variable—called USED60—representing whether the person used fewer or more than 60 tissues. (The values of USED60 in the data set are “LESS” and “MORE.”) A MINITAB analysis of this variable yielded the printout displayed in Figure SIA6.2 on p. 343.

MINITAB test of $p = .5$ $p = .5$ for KLEENEX survey

On the MINITAB printout, x represents the number of the 250 people with colds that used more than 60 tissues. Note that $x = 154.$ $x = 154.$ This value is used to compute the test statistic $z = 3.67,$ $z = 3.67,$ highlighted on the printout. The p-value of the test, also highlighted on the printout, is $p - v a l u e = .000 .$ $p - v a l u e = .000 .$ Since this value is less than $α = .05,$ $α = .05,$ there is sufficient evidence (at $α = .05$ $α = .05$ ) to reject $H_{0};$ $H_{0};$ we conclude that the proportion of all KLEENEX^® users who use more than 60 tissues when they have a cold exceeds .5. This conclusion again supports, the company’s decision to put more than 60 tissues in an anti-viral box of KLEENEX.

Data Set: TISSUES

Exercises 6.73–6.93

Understanding the Principles

6.73 What type of data, quantitative or qualitative, is typically associated with making inferences about a population proportion p?
6.74 What conditions are required for a valid large-sample test for p?

Learning the Mechanics

6.75 For the binomial sample sizes and null-hypothesized values of p in each part, determine whether the sample size is large enough to use the normal approximation methodology presented in this section to conduct a test of the null hypothesis $H_{0} : p = p_{0} .$ $H_{0} : p = p_{0} .$
1. $n = 500, p_{0} = .05$ $n = 500, p_{0} = .05$
2. $n = 100, p_{0} = .99$ $n = 100, p_{0} = .99$
3. $n = 50, p_{0} = .2$ $n = 50, p_{0} = .2$
4. $n = 20, p_{0} = .2$ $n = 20, p_{0} = .2$
5. $n = 10, p_{0} = .4$ $n = 10, p_{0} = .4$
6.76 Suppose a random sample of 100 observations from a binomial population gives a value of $\hat{p} = .69$ $\hat{p} = .69$ and you wish to test the null hypothesis that the population parameter p is equal to .75 against the alternative hypothesis that p is less than .75.
1. Noting that $\hat{p} = .69,$ $\hat{p} = .69,$ what does your intuition tell you? Does the value of $\hat{p}$ $\hat{p}$ appear to contradict the null hypothesis?
2. Use the large-sample z-test to test $H_{0} : p = .75$ $H_{0} : p = .75$ against the alternative hypothesis $H_{a} : p < .75 .$ $H_{a} : p < .75 .$ Use $α = .05 .$ $α = .05 .$ How do the test results compare with your intuitive decision from part a?
3. Find and interpret the observed significance level of the test you conducted in part b.
6.77 Suppose the sample in Exercise 6.76 has produced $\hat{p} = .84$ $\hat{p} = .84$ and we wish to test $H_{0} : p = .9$ $H_{0} : p = .9$ against the alternative $H_{a} : p < .9 .$ $H_{a} : p < .9 .$
1. Calculate the value of the z statistic for this test.
2. Note that the numerator of the z statistic $(\hat{p} - p_{0} = .84 - .90 = - .06)$ $(\hat{p} - p_{0} = .84 - .90 = - .06)$ is the same as for Exercise 6.76. Considering this, why is the absolute value of z for this exercise larger than that calculated in Exercise 6.76?
3. Complete the test, using $α = .05,$ $α = .05,$ and interpret the result.
4. Find the observed significance level for the test and interpret its value.
6.78 A random sample of 100 observations is selected from a binomial population with unknown probability of success p. The computed value of $\hat{p}$ $\hat{p}$ is equal to .74.
1. Test $H_{0} : p = .65$ $H_{0} : p = .65$ against $H_{a} : p > .65 .$ $H_{a} : p > .65 .$ Use $α = .01 .$ $α = .01 .$
2. Test $H_{0} : p = .65$ $H_{0} : p = .65$ against $H_{a} : p > .65 .$ $H_{a} : p > .65 .$ Use $α = .10 .$ $α = .10 .$
3. Test $H_{0} : p = .90$ $H_{0} : p = .90$ against $H_{a} : p \neq .90 .$ $H_{a} : p \neq .90 .$ Use $α = .05 .$ $α = .05 .$
4. Form a 95% confidence interval for p.
5. Form a 99% confidence interval for p.
SNACK 6.79 Refer to Exercise 5.58 (p. 280), in which 50Fifty consumers taste-tested a new snack food.
1. Test $H_{0} : p = .5$ $H_{0} : p = .5$ against $H_{a} : p > .5,$ $H_{a} : p > .5,$ where p is the proportion of customers who do not like the snack food. Use $α = .10 .$ $α = .10 .$
2. Report the observed significance level of your test.

Applet Exercise 6.5

Use the applet entitled Hypotheses Test for a Proportion to investigate the relationships between the probabilities of Type I and Type II errors occurring at levels .05 and .01. For this exercise, use $n = 100,$ $n = 100,$ true $p = 0.5,$ $p = 0.5,$ and alternative not equal.

Set null $p = .5 .$ $p = .5 .$ What happens to the proportion of times the null hypothesis is rejected at the .05 level and at the .01 level as the applet is run more and more times? What type of error has occurred when the null hypothesis is rejected in this situation? Based on your results, is this type of error more likely to occur at level .05 or at level .01? Explain.
Set null $p = .6 .$ $p = .6 .$ What happens to the proportion of times the null hypothesis is not rejected at the .05 level and at the .01 level as the applet is run more and more times? What type of error has occurred when the null hypothesis is not rejected in this situation? Based on your results, is this type of error more likely to occur at level .05 or at level .01? Explain.
Use your results from parts a and b to make a general statement about the probabilities of Type I and Type II errors at levels .05 and .01.

Applet Exercise 6.6

Use the applet entitled Hypotheses Test for a Proportion to investigate the effect of the true population proportion p on the probability of a Type I error occurring. For this exercise, use $n = 100$ $n = 100$ and alternative not equal.

Set true $p = .5$ $p = .5$ and null $p = .5 .$ $p = .5 .$ Run the applet several times, and record the proportion of times the null hypothesis is rejected at the .01 level.
Clear the applet and repeat part a for true $p = .1$ $p = .1$ and null $p = .1 .$ $p = .1 .$ Then repeat one more time for true $p = .01$ $p = .01$ and null $p = .01 .$ $p = .01 .$
Based on your results from parts a and b, what can you conclude about the probability of a Type I error occurring as the true population proportion gets closer to 0?

Applying the Concepts—Basic

MUSIC 6.80 Paying for music downloads. If you use the Internet, have you ever paid to access or download music? This was one of the questions of interest in a recent Pew Internet and American Life Project Survey (Oct. 2010). In a representative sample of 755 adults who use the Internet, 506 stated that they have paid to download music. Let p represent the true proportion of all Internet-using adults who have paid to download music.
1. Compute a point estimate of p.
2. Set up the null and alternative hypotheses for testing whether the true proportion of all Internet-using adults who have paid to download music exceeds .7.
3. Compute the test statistic for part b.
4. Find the rejection region for the test if $α = .01$ $α = .01$ .
5. Find the p-value for the test.
6. Make the appropriate conclusion using the rejection region.
7. Make the appropriate conclusion using the p-value.
6.81 Underwater sound-locating ability of alligators. Alligators have shown the ability to determine the direction of an airborne sound. But can they locate underwater sounds? This was the subject of research published in the Journal of Herpetology (Dec. 2014). Alligators inhabiting the flood control canals in the Florida Everglades were monitored for movement toward a sound produced from a submerged diving bell. Movements within a 180° arc of the direction toward the sound were scored as movements toward the sound; all movements in other directions were scored as movements away from the sound. Consequently, the researchers assumed that the proportion of movements toward the sound expected by chance is $180 ° / 360 ° = .5$ $180 ° / 360 ° = .5$ . In a sample of $n = 50$ $n = 50$ alligators, 42 moved toward the underwater sound.
1. Give the null and alternative hypotheses for testing whether the true proportion of alligators that move toward the underwater sound is higher than expected by chance.
2. In a sample of $n = 50$ $n = 50$ alligators, assume that 42 moved toward the underwater sound. Use this information to compute an estimate of the true proportion of alligators that move toward the underwater sound.
3. Compute the test statistic for this study.
4. Compute the observed significance level (p-value) of the test.
5. Make the appropriate conclusion in the words of the problem.
6.82 Gummy bears: red or yellow? Chance (Winter 2010) presented a lesson in hypothesis testing carried out by medical students in a biostatistics class. Students were blind-folded and then given a red-colored or yellow-colored gummy bear to chew. (Half the students were randomly assigned to receive the red gummy bear and half to receive the yellow bear. The students could not see what color gummy bear they were given.) After chewing, the students were asked to guess the color of the candy based on the flavor. Of the 121 students who participated in the study, 97 correctly identified the color of the gummy bear.
1. If there is no relationship between color and gummy bear flavor, what proportion of the population of students will correctly identify the color?
2. Specify the null and alternative hypotheses for testing whether color and flavor are related.
3. Carry out the test and give the appropriate conclusion at $α = .01$ $α = .01$ . Use the p-value of the test to make your decision.
6.83 Dehorning of dairy calves. For safety reasons, calf dehorning has become a routine practice at dairy farms. A report by Europe’s Standing Committee on the Food Chain and Animal Health (SANKO) stated that 80% of European dairy farms carry out calf dehorning. A later study, published in the Journal of Dairy Science (Vol. 94, 2011), found that in a sample of 639 Italian dairy farms, 515 dehorn calves. Does the Journal of Dairy Science study support or refute the figure reported by SANKO? Explain.
6.84 Teenagers’ use of emoticons in school writing. Refer to the Pew Internet and American Life Project (Apr. 2008) survey of the writing habits of U.S. teenagers, Exercise 5.69 (p. 343). Recall that in a random sample of 700 teenagers, 448 admitted to using at least one informal element in school writing assignments. [Note: Emoticons, such as using the symbol “:)” to represent a smile, and abbreviations, such as writing “LOL” for “laughing out loud,” are considered informal elements.] Is there evidence to indicate that less than 65% of all U.S. teenagers have used at least one informal element in school writing assignments? Test the relevant hypotheses using $α = .05 .$ $α = .05 .$
6.85 TV subscription streaming. “Streaming” of television programs is trending upward. According to The Harris Poll (Aug. 26, 2013), over one-third of Americans qualify as “subscription streamers,” i.e., those who watch streamed TV programs through a subscription service such as Netflix, Hulu Plus, or Amazon Prime. The poll included 2,242 adult TV viewers, of which 785 are subscription streamers. Based on this result, can you conclude that the true fraction of adult TV viewers who are subscription streamers differs from one-third? Carry out the test using a Type I error rate of .10.

Applying the Concepts—Intermediate

6.86 Underwater acoustic communication. Refer to the IEEE Journal of Oceanic Engineering (Apr. 2013) study of the characteristics of subcarriers—telecommunication signals carried on top of another—for underwater acoustic communications, Exercise 4.68 (p. 194). Recall that aA subcarrier can be classified as either a data subcarrier (used for data transmissions), a pilot subcarrier (used for channel estimation and synchronization), or a null subcarrier (used for direct current and guard banks transmitting no signal). In a sample of 1,024 subcarrier signals transmitted off the coast of Martha’s Vineyard, 672 were determined to be data subcarriers, 256 pilot subcarriers, and 96 null subcarriers. Suppose a communications engineer who works near Martha’s Vineyard believes that fewer than 70% of all subcarrier signals transmitted in the area are data subcarriers. Is there evidence to support this belief? Test using $α = .05$ $α = .05$ .
6.87 Identifying organisms using a computer. National Science Education Standards recommend that all life science students be exposed to methods of identifying unknown biological specimens. Due to certain limitations of traditional identification methods, biology professors at Slippery Rock University (SRU) developed a computer-aided system for identifying common conifers (deciduous trees) called Confir ID (The American Biology Teacher, May 2010). A sample of 171 life science students were exposed to both a traditional method of identifying conifers and Confir ID and then asked which method they preferred. As a result, 138 students indicated their preference for Confir ID. In order to change the life sciences curriculum at SRU to include Confir ID, the biology department requires that more than 70% of the students prefer the new, computerized method. Should Confir ID be added to the curriculum at SRU? Explain your reasoning.
6.88 Organic certified coffee. Coffee markets that conform to organic standards focus on the environmental aspect of coffee growing, such as the use of shade trees and a reduced reliance on chemical pesticides. A study of organic coffee growers was published in Food Policy (Vol. 36, 2010). In a representative sample of 845 coffee growers from southern Mexico, 417 growers were certified to sell to organic coffee markets while 77 growers were transitioning to become organic certified. In the United States, 60% of the coffee growers are organic certified. Is there evidence to indicate that fewer than 60% of the coffee growers in southern Mexico are either organic certified or transitioning to become certified? State your conclusion so that there is only a 5% chance of making a Type I error.
6.89 Detection of motorcycles while driving. The factors that affect the visibility of powered two-wheelers (PTWs)—motorcycles—on the road were investigated in Accident Analysis and Prevention (Vol. 44, 2012). A visual search study was conducted in which viewers were presented with pictures of driving scenarios and asked to identify the presence or absence of a PTW. Of interest to the researchers is the detection rate, i.e., the proportion of pictures showing a PTW in which the viewer actually detected the presence of the PTW. Suppose that, in theory, the true detection rate for pictures of PTWs is .70. The study revealed that in a sample of 2,376 pictures that included a PTW, only 1,554 were detected by the viewers. Use this result to test the theory at $α = .10$ $α = .10$
DINDEX 6.90 Defensibility of a landscape. Settlers often chose sites to live and build communities in based on the landscape’s defensibility. The defensibility of archaeological sites in the Northwest Pacific was investigated in the Journal of Anthropological Archaeology (Vol. 35, 2014). A defensibility index was calculated for each in a sample of 23 sites that included a trench embankment. These index values (on a scale of 1 to 100) are listed in the accompanying table. The researcher determined that the mean defensibility index of the entire Northwest Pacific landscape is 47. Of interest is whether the defensibility index of a sampled site exceeds this mean.

Alternate View

70 70 68 67 67 66 65 63 63 59 57 57

57 55 55 54 52 52 51 46 46 44 37

Source: Bocinsky, R. K. “Extrinsic site defensibility and landscape-based archaeological inference: An example from the Northwest Coast.” Journal of Anthropological Archaeology, Vol. 35, 2014 (Table 3).

SAS Output for Exercise 6.90
1. Estimate p, the true proportion of all trench embankment sites that have a defensibility index that exceeds the mean of the entire Northwest Pacific landscape.
2. An archaeologist wants to conduct a statistical test of hypothesis to determine if p is larger than .60. Give the null and alternative hypotheses for this test.
3. Demonstrate that the sample is not large enough to apply the large-sample test of a binomial proportion.
4. The SAS printout shown on p. 407 gives the results of an exact binomial proportion test of the hypotheses, part b. Locate the p-value of the test (bottom of the printout) and interpret the results.
6.91 Astronomy students and the Big Bang Theory. Indiana University professors investigated first-year college students’ knowledge of astronomy (Astronomy Education Review, Vol. 2, 2003). One concept of interest was the Big Bang theory of the creation of the universe. In a sample of 148 freshman students, 37 believed that the Big Bang theory accurately described the creation of planetary systems. Based on this information, would you be willing to state that more than 20% of all freshman college students believe in the Big Bang theory? How confident are you of your decision?

Applying the Concepts—Advanced

6.92 Killing insects with low oxygen. A group of Australian entomological toxicologists investigated the impact of exposure to low oxygen on the mortality of insects (Journal of Agricultural, Biological, and Environmental Statistics, Sept. 2000). Thousands of adult rice weevils were placed in a chamber filled with wheat grain, and the chamber was exposed to nitrogen gas for 4 days. Insects were assessed as dead or alive 24 hours after exposure. At the conclusion of the experiment, 31,386 weevils were dead and 35 weevils were found alive. Previous studies had shown a 99% mortality rate in adult rice weevils exposed to carbon dioxide for 4 days. What advice would you give to the entomologists regarding the use of nitrogen or carbon dioxide for killing rice weevils?
6.93 Male fetal deaths following 9/11/2001. According to research published in BMC Public Health (May 25, 2010), the terrorist attacks of September 11, 2001, may have led to a spike in miscarriages of male babies. Researchers from the University of California, using data from the National Vital Statistics System, reported that the male fetal death rate in September of that year was significantly higher than expected. Suppose it is known that the miscarriage rate for pregnant women expecting a boy in the United States is 3 in 1,000. In a random sample of 2,000 pregnant women expecting a baby boy during September of 2001, how many would need to have experienced a fetal death (miscarriage) in order to support the claim (at $α = .05$ $α = .05$ ) that the male fetal death rate in September 2001 was higher than expected?

..................Content has been hidden....................

You can't read the all page of ebook, please click here login for view all page.

Table of Contents for 6.6 Large-Sample Test of Hypothesis about a Population Proportion

Create new playlist

Sign In

Sign Up

Table of Contents for
6.6 Large-Sample Test of Hypothesis about a Population Proportion