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The article referenced in Exercise \(10.34\) also reported that 470 of 1000 randomly selected adult Americans thought that the quality of movies being produced was getting worse. a. Is there convincing evidence that fewer than half of adult Americans believe that movie quality is getting worse? Use a significance level of \(.05\). b. Suppose that the sample size had been 100 instead of 1000 , and that 47 thought that the movie quality was getting worse (so that the sample proportion is still . 47 ). Based on this sample of 100 , is there convincing evidence that fewer than half of adult Americans believe that movie quality is getting worse? Use a significance level of \(.05\). c. Write a few sentences explaining why different conclusions were reached in the hypothesis tests of Parts (a) and (b).

Short Answer

Expert verified
The difference in conclusions can be attributed to the difference in the sample size. Even though the sample proportion is the same in both cases, the larger sample size in the first case inflates the Z value and therefore the P-value, leading to a different conclusion about the population proportion.

Step by step solution

01

Hypothesis Formulation

First, we need to formulate our null hypothesis \(H_0\) and alternative hypothesis \(H_A\). Since the question is asking if there is convincing evidence that fewer than half of adult Americans believe that movie quality is getting worse, the null hypothesis is \(H_0: p = 0.5\) and the alternative hypothesis is \(H_A: p < 0.5\).
02

Test Statistics Calculation for Sample Size 1000

Next, we calculate the test statistics which is given by the formula \(Z = \frac{p - P}{\sqrt{P(1-P)/n}}\). Here, \(p\) is our sample proportion which is \(.47\), \(P\) is our population proportion under \(H_0\) which is \(.5\), and \(n\) is our sample size which is 1000. After substituting these values, the Z score can be calculated.
03

P-value Calculation and Conclusion for Sample Size 1000

Once we have the Z score, we calculate the P-value which is the probability of getting a test statistic as extreme or more extreme than what was observed under the null hypothesis. We calculate this using standard normal distribution tables or calculator. If P-value is less than the significance level, we reject the null hypothesis and state that there is convincing evidence that fewer than half of adults believe the movie quality is getting worse.
04

Test Statistics Calculation for Sample Size 100

We repeat steps 2 and 3 for a sample size of 100 and sample proportion of .47. Note that when repeating these steps the numerator in the test statistics calculation formula (which is the difference between the sample and population proportions) will remain the same while the denominator (which is the standard error) will increase due to smaller sample size.
05

P-value Calculation and Conclusion for Sample Size 100

Again calculate the P-value and if it is less than the significance level, reject the null hypothesis and state that there is convincing evidence that fewer than half of adults believe the movie quality is getting worse.
06

Comparison of Results

Lastly, compare the results of hypothesis tests for both samples. The difference in conclusions can mainly be attributed to the difference in sample sizes. Larger sample sizes reduce the standard error, making the test statistic larger for a given observed difference between sample and population proportions.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Null and Alternative Hypotheses
Understanding the null and alternative hypotheses is crucial for hypothesis testing in statistics. The null hypothesis (\( H_0 \)) represents the default claim—what we assume to be true before collecting any data. For instance, in the context of our exercise, the null hypothesis is that exactly half of the adult Americans (\( p = 0.5 \) think that movie quality is getting worse. On the other hand, the alternative hypothesis (\( H_A \) or sometimes denoted as (\( H_1 \) presents a competing claim we are trying to find evidence for; it contradicts the null hypothesis. In our example, the alternative hypothesis is that fewer than half of the adult Americans (\( p < 0.5 \) believe movie quality is deteriorating.

The function of these hypotheses is not to prove anything definitively but to set the stage for statistical testing. By establishing these two opposing hypotheses, we create a scenario that allows us to use statistical evidence to either reject the null hypothesis in favor of the alternative or fail to reject the null hypothesis. This methodical approach prevents us from jumping to conclusions based on subjective or anecdotal evidence.
Test Statistics Calculation
Once hypotheses are defined, the next step is to calculate a test statistic. This measurement tells us how far our sample statistic is from the parameter stated in the null hypothesis, measured in terms of standard error. The formula is dependent on the type of test being conducted, but generally, it's a ratio: the difference between the sample statistic and the null hypothesis value divided by the standard error of the statistic.

In our exercise, we use a Z-test statistic formula for proportions: \[ Z = \frac{p - P}{\sqrt{P(1-P)/n}} \]Here, \(p\) is the sample proportion which represents the proportion of individuals in our sample who believe the quality of movies is worsening. \(P\) stands for the population proportion under the null hypothesis, and \(n\) is our sample size. The larger the value of \(Z\), the further our sample proportion is from the null hypothesis. Calculating this value helps us understand whether any observed differences are statistically significant or just due to random chance.
P-value and Significance Level
After calculating the test statistics, we determine the p-value. The p-value is the probability of observing a test statistic as extreme or more extreme than the one calculated, assuming the null hypothesis is true. It quantifies the strength of the evidence against the null hypothesis. The significance level (\( \alpha \)), is a predetermined threshold for making this decision—commonly set at 0.05.

If the p-value is less than or equal to the significance level, we reject the null hypothesis, concluding that the evidence suggests a statistically significant effect. Otherwise, we fail to reject the null hypothesis, implying that the evidence isn't strong enough to warrant a conclusion that there's an effect. It's important to realize that a p-value does not measure the size of an effect or the importance of a result, but rather the probability of the result occurring by chance.
Statistical Hypothesis Comparison
Comparing statistical hypotheses means evaluating the strength of the evidence for or against our null hypothesis. In hypothesis testing, we're dealing with two potential outcomes for our p-value: either we have sufficient evidence to reject the null hypothesis, or we do not. When comparing different hypothesis tests, as seen in the variation of sample sizes in our exercise, conclusions can differ due to changes in sample size, variability, and effect size.

A larger sample size often means a smaller standard error, which can lead to a larger Z value for the same observed effect, possibly resulting in a smaller p-value. This was demonstrated in our exercise where different conclusions were reached with sample sizes of 1000 and 100, respectively. It's vital for students to understand that while the process of hypothesis testing is the same, outcomes can change with different sample sizes, which can significantly affect the power of the test and the confidence in our conclusions.

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Most popular questions from this chapter

When a published article reports the results of many hypothesis tests, the \(P\) -values are not usually given. Instead, the following type of coding scheme is frequently used: \({ }^{*} p=.05,{ }^{* *} p=.01,{ }^{* * *} p=.001,{ }^{* * * *} p=.0001\). Which of the symbols would be used to code for each of the following \(P\) -values? a. \(.037\) c. 072 b. \(.0026\) d. \(.0003\)

The report "2005 Electronic Monitoring \& Surveillance Survey: Many Companies Monitoring, Recording, Videotaping-and Firing-Employees" (American Management Association, 2005) summarized the results of a survey of 526 U.S. businesses. Four hundred of these companies indicated that they monitor employees' web site visits. For purposes of this exercise, assume that it is reasonable to regard this sample as representative of businesses in the United States. a. Is there sufficient evidence to conclude that more than \(75 \%\) of U.S. businesses monitor employees' web site visits? Test the appropriate hypotheses using a significance level of \(.01\). b. Is there sufficient evidence to conclude that a majority of U.S. businesses monitor employees' web site visits? Test the appropriate hypotheses using a significance level of \(.01\).

A well-designed and safe workplace can contribute greatly to increasing productivity. It is especially important that workers not be asked to perform tasks, such as lifting, that exceed their capabilities. The following data on maximum weight of lift (MWOL, in kilograms) for a frequency of 4 lifts per minute were reported in the article "The Effects of Speed, Frequency, and Load on Measured Hand Forces for a Floor-to-Knuckle Lifting Task" (Ergonomics \([1992]: 833-843)\) : \(\begin{array}{lllll}25.8 & 36.6 & 26.3 & 21.8 & 27.2\end{array}\) Suppose that it is reasonable to regard the sample as a random sample from the population of healthy males, age \(18-30\). Do the data suggest that the population mean MWOL exceeds 25 ? Carry out a test of the relevant hypotheses using a \(.05\) significance level.

Typically, only very brave students are willing to speak out in a college classroom. Student participation may be especially difficult if the individual is from a different culture or country. The article "An Assessment of Class Participation by International Graduate Students" \((\) Journal of College Student Development \([1995]: 132-\) 140) considered a numerical "speaking-up" scale, with possible values from 3 to 15 (a low value means that a student rarely speaks). For a random sample of 64 males from Asian countries where English is not the official language, the sample mean and sample standard deviation were \(8.75\) and \(2.57\), respectively. Suppose that the mean for the population of all males having English as their native language is \(10.0\) (suggested by data in the article). Does it appear that the population mean for males from non-English-speaking Asian countries is smaller than \(10.0 ?\)

According to the article "Workaholism in Organizations: Gender Differences" (Sex Roles [1999]: \(333-346\) ), the following data were reported on 1996 income for random samples of male and female MBA graduates from a certain Canadian business school: \begin{tabular}{lccc} & \(\boldsymbol{N}\) & \(\overline{\boldsymbol{x}}\) & \(\boldsymbol{s}\) \\ \hline Males & 258 & \(\$ 133,442\) & \(\$ 131,090\) \\ Females & 233 & \(\$ 105,156\) & \(\$ 98,525\) \\ \hline \end{tabular} Note: These salary figures are in Canadian dollars. a. Test the hypothesis that the mean salary of male MBA graduates from this school was in excess of \(\$ 100,000\) in \(1996 .\) b. Is there convincing evidence that the mean salary for all female MBA graduates is above \(\$ 100,000 ?\) Test using \(\alpha=.10\) c. If a significance level of \(.05\) or \(.01\) were used instead of \(.10\) in the test of Part (b), would you still reach the same conclusion? Explain.

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