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Chapter 9: Problem 29

Generally speaking, would you say that most people can be trusted? A random sample of \(n_{1}=250\) people in Chicago ages \(18-25\) showed that \(r_{1}=45\) said yes. Another random sample of \(n_{2}=280\) people in Chicago ages \(35-45\) showed that \(r_{2}=71\) said yes (based on information from the National Opinion Research Center, University of Chicago). Does this indicate that the population proportion of trusting people in Chicago is higher for the older group? Use \(\alpha=0.05\).

Short Answer

Expert verified
The data suggests that the proportion of trusting people in the older age group is significantly higher.

Step by step solution

01

Define Hypotheses

We define our null hypothesis \(H_0\) and alternative hypothesis \(H_1\). \(H_0: p_1 \geq p_2\) (the proportion of trusting people in the younger group is greater than or equal to that in the older group) and \(H_1: p_1 < p_2\) (the proportion of trusting people in the younger group is less than that in the older group).
02

Calculate Sample Proportions

Calculate the sample proportions for each group. \(\hat{p}_1 = \frac{r_1}{n_1} = \frac{45}{250} = 0.18\) for the 18-25 age group, and \(\hat{p}_2 = \frac{r_2}{n_2} = \frac{71}{280} \approx 0.2536\) for the 35-45 age group.
03

Pooled Sample Proportion

Compute the pooled sample proportion \(\hat{p}\), which combines the information of both samples: \(\hat{p} = \frac{r_1 + r_2}{n_1 + n_2} = \frac{45 + 71}{250 + 280} = \frac{116}{530} \approx 0.2189\).
04

Standard Error Calculation

Calculate the standard error (SE) for the difference in proportions: \[SE = \sqrt{\hat{p}(1 - \hat{p})\left(\frac{1}{n_1} + \frac{1}{n_2}\right)} = \sqrt{0.2189 \times 0.7811 \times \left(\frac{1}{250} + \frac{1}{280}\right)} \approx 0.0397\].
05

Z-Score Calculation

Calculate the Z-score for the difference in sample proportions. \[Z = \frac{\hat{p}_1 - \hat{p}_2}{SE} = \frac{0.18 - 0.2536}{0.0397} \approx -1.858\].
06

Decision Rule and Conclusion

The critical value for a one-tailed test with \(\alpha = 0.05\) is approximately -1.645. Since the calculated Z-score \(-1.858\) is less than \(-1.645\), we reject the null hypothesis. This suggests that there is significant evidence to conclude that the proportion of trusting people is higher in the older age group.

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Key Concepts

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

Population Proportion
Understanding population proportion is essential when we want to analyze preferences or behaviors across a large group based on sample data. It refers to the ratio of individuals in a population who possess a certain characteristic.
For example, in our exercise, we want to know the population proportion of trusting people in different age groups in Chicago.
  • The population proportion helps us to draw conclusions about the entire population, based on the sample.
  • When the population is large, it's often impractical to survey everyone. Hence, we rely on a smaller sample.
  • This concept is critical in making inferences in hypothesis testing.
Therefore, population proportion is the backbone for statistical investigations that help in decision-making processes.
Sample Proportion
While population proportion covers the whole group, sample proportion focuses on a smaller subset of that group. It鈥檚 calculated by taking the number of favorable outcomes and dividing it by the total number of observations in the sample.
In the exercise, sample proportions were calculated as:
  • For the 18-25 age group: \[\hat{p}_1 = \frac{45}{250} = 0.18\]
  • For the 35-45 age group:\[\hat{p}_2 = \frac{71}{280} \approx 0.2536\]
This measure tells us the fraction of the sample that exhibits the characteristic of interest. By comparing sample proportions from different groups, like our age-based groups, we identify significant differences or similarities.
Z-score
To understand how much a sample proportion deviates from the population proportion, we use the Z-score. It's a statistical measure that tells us how many standard deviations an element is from the mean.

In hypothesis testing, the Z-score enables us to determine the probability of observing our sample statistic under the null hypothesis. In the exercise, it was calculated as:\[Z = \frac{\hat{p}_1 - \hat{p}_2}{SE} \approx -1.858\]
  • The Z-score helps us decide whether the observed difference is due to random chance or a true difference in population proportions.
  • If the Z-score is beyond a critical value, we reject the null hypothesis, signaling that the sample statistic is not consistent with the null hypothesis.
Standard Error
The standard error (SE) is a statistical term that measures the accuracy with which a sample represents a population. For proportions, it helps approximate the variability or dispersion of sample proportion predictions from the overall population proportion.
In this exercise, the standard error was calculated using:\[SE = \sqrt{\hat{p}(1 - \hat{p})\left(\frac{1}{n_1} + \frac{1}{n_2}\right)} \approx 0.0397\]
  • SE reflects how much the sample statistic may differ from the actual population parameter.
  • A smaller SE indicates that the sample proportion is a more precise estimate of the population proportion.
  • It's used along with the Z-score to assess the validity of our hypothesis testing conclusions.
One-tailed Test
Hypothesis tests can be one-tailed or two-tailed. A one-tailed test assesses if a parameter is either greater than or less than a certain value, but not both.
In this case, the test examines if the population proportion of trusting people in the older age group is greater than that in the younger age group.
  • One-tailed tests increase the power to detect an effect if it exists in the predicted direction.
  • In our exercise, since \[Z = -1.858\] was less than the critical value of \[-1.645\], the null hypothesis was rejected, favoring the alternative hypothesis that more older individuals are trusting.
  • This approach simplifies hypothesis testing by focusing on a specific direction of change or difference.
Thus, when you have a directional hypothesis, the one-tailed test offers a focused method to validate or refute your hypothesis.

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Most popular questions from this chapter

In general, if sample data are such that the null hypothesis is rejected at the \(\alpha=1 \%\) level of significance based on a two-tailed test, is \(H_{0}\) also rejected at the \(\alpha=1 \%\) level of significance for a corresponding one-tailed test? Explain.

Would you favor spending more federal tax money on the arts? This question was asked by a research group on behalf of The National Institute (Reference: Painting by Numbers, J. Wypijewski, University of California Press). Of a random sample of \(n_{1}=93\) politically conservative voters, \(r_{1}=21\) responded yes. Another random sample of \(n_{2}=83\) politically moderate voters showed that \(r_{2}=22\) responded yes. Does this information indicate that the population proportion of conservative voters inclined to spend more federal tax money on funding the arts is less than the proportion of moderate voters so inclined? Use \(\alpha=0.05\).

A study of fox rabies in southern Germany gave the following information about different regions and the occurrence of rabies in each region (Reference: B. Sayers, et al., "A Pattern Analysis Study of a Wildlife Rabies Epizootic," Medical Informatics 2:11-34). Based on information from this article, a random sample of \(n_{1}=16\) locations in region I gave the following information about the number of cases of fox rabies near that location. \(\begin{array}{llllllll}x_{1} \text { : Region I data } & 1 & 8 & 8 & 8 & 7 & 8 & 8 \\ & 3 & 3 & 3 & 2 & 5 & 1 & 4\end{array}\) A second random sample of \(n_{2}=15\) locations in region II gave the following information about the number of cases of fox rabies near that location. \(x_{2}\) : Region II data \(\quad 1\) i. Use a calculator with sample mean and sample standard deviation keys to verify that \(\bar{x}_{1}=4.75\) with \(s_{1} \approx 2.82\) in region \(\mathrm{I}\) and \(\bar{x}_{2} \approx 3.93\) with \(s_{2} \approx 2.43\) in region II. ii. Does this information indicate that there is a difference (either way) in the mean number of cases of fox rabies between the two regions? Use a \(5 \%\) level of significance. (Assume the distribution of rabies cases in both regions is mound-shaped and approximately normal.)

When testing the difference of means for paired data, what is the null hypothesis?

In environmental studies, sex ratios are of great importance. Wolf society, packs, and ecology have been studied extensively at different locations in the U.S. and foreign countries. Sex ratios for eight study sites in northern Europe are shown below (based on The Wolf by L. D. Mech, University of Minnesota Press). \(\begin{array}{lcc} \hline \text { Location of Wolf Pack } & \text { \% Males (Winter) } & \text { \% Males (Summer) } \\ \hline \text { Finland } & 72 & 53 \\ \text { Finland } & 47 & 51 \\ \text { Finland } & 89 & 72 \\ \text { Lapland } & 55 & 48 \\ \text { Lapland } & 64 & 55 \\ \text { Russia } & 50 & 50 \\ \text { Russia } & 41 & 50 \\ \text { Russia } & 55 & 45 \\ \hline \end{array}\) It is hypothesized that in winter, "loner" males (not present in summer packs) join the pack to increase survival rate. Use a \(5 \%\) level of significance to test the claim that the average percentage of males in a wolf pack is higher in winter.
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