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Chapter 10: Problem 90

Is the response rate for questionnaires affected by including some sort of incentive to respond along with the questionnaire? In one experiment, 110 questionnaires with no incentive resulted in 75 being retumed, whereas 98 questionnaires that included a chance to win a lottery yielded 66 responses ("Charities, No; Lotteries, No; Cash, Yes," Public Opinion Q., 1996: 542-562). Does this data suggest that including an incentive increases the likelihood of a response? State and test the relevant hypotheses at significance level \(.10\) by using the \(P\)-value method.

Short Answer

Expert verified
Incentives do not significantly increase response rates at a 0.10 significance level.

Step by step solution

01

Define Hypotheses

First, establish the null and alternative hypotheses. Let \( p_1 \) be the response rate without an incentive, and \( p_2 \) be the response rate with an incentive. - Null Hypothesis \( (H_0) \): \( p_1 = p_2 \) (no effect of incentive).- Alternative Hypothesis \( (H_a) \): \( p_1 < p_2 \) (incentive increases response rate).
02

Collect Sample Data

Extract the sample data from the problem: - Without incentive: \( n_1 = 110 \), \( x_1 = 75 \).- With incentive: \( n_2 = 98 \), \( x_2 = 66 \). Calculate the sample proportions: - \( \hat{p}_1 = \frac{75}{110} \approx 0.6818 \). - \( \hat{p}_2 = \frac{66}{98} \approx 0.6735 \).
03

Calculate Standard Error

Calculate the pooled standard error for \( \hat{p}_1 - \hat{p}_2 \): \[ \hat{p} = \frac{x_1 + x_2}{n_1 + n_2} = \frac{75 + 66}{110 + 98} = \frac{141}{208} \approx 0.6784 \] \[ SE = \sqrt{\hat{p}(1 - \hat{p}) \left( \frac{1}{n_1} + \frac{1}{n_2} \right)} = \sqrt{0.6784 \times 0.3216 \left( \frac{1}{110} + \frac{1}{98} \right)} \approx 0.065 \]
04

Test Statistic Calculation

Calculate the test statistic using the formula: \[ z = \frac{\hat{p}_1 - \hat{p}_2}{SE} = \frac{0.6818 - 0.6735}{0.065} \approx 0.128 \]
05

Determine P-Value

Look up the P-value for \( z = 0.128 \) using a standard normal distribution table. The P-value \( \approx 0.4483 \), since it is much larger than \( 0.10 \), indicating that the result is not statistically significant.
06

Conclusion

Compare the P-value to the significance level \( \alpha = 0.10 \). Since \( \text{P-value} \approx 0.4483 \) is greater than \( 0.10 \), we fail to reject the null hypothesis. There is not enough evidence to conclude that including an incentive increases the response rate.

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Key Concepts

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

P-value Method
In hypothesis testing, the P-value method is a popular approach used to determine the significance of the test results. The P-value represents the probability of observing a test statistic as extreme as, or more extreme than, the observed results, assuming that the null hypothesis is true.

To utilize the P-value method, perform the following steps:
  • Compute the test statistic, which measures how far the sample statistic is from the null hypothesis parameter in terms of standard error.
  • Reference a statistical table or software to find the P-value corresponding to the calculated test statistic.
  • Compare the P-value to a pre-determined significance level, often denoted as \( \alpha \), such as 0.05 or 0.10, depending on the context of the study.
If the P-value is smaller than \( \alpha \), you reject the null hypothesis, suggesting that the observed data are statistically significant. However, if the P-value is larger, you do not have enough evidence to reject the null hypothesis. In our exercise, since the P-value was 0.4483, which is greater than \( \alpha = 0.10 \), this led to the conclusion of failing to reject the null hypothesis.
Null and Alternative Hypotheses
When conducting hypothesis testing, first define the null and alternative hypotheses. These hypotheses are contrasting statements about the population parameter you are investigating.

  • The null hypothesis, denoted as \( H_0 \), generally reflects the status quo or no effect. For this problem, the null hypothesis states that there is no difference in the response rates of questionnaires with or without incentives, meaning \( p_1 = p_2 \).
  • The alternative hypothesis, denoted as \( H_a \), reflects the effect or difference you are testing. Here, it posits that the response rate with incentives is higher than without, depicted as \( p_1 < p_2 \).
Identifying these hypotheses correctly is crucial because they provide the framework for the statistical test. They help determine the direction of the test and influence the calculation of the P-value.
Sample Proportion
Sample proportion is an essential concept in statistics, representing the ratio of favorable outcomes in a sample to the total sample size. It gives us an estimate of the population proportion and is crucial for hypothesis testing.

In the given problem, the sample proportions were calculated for both groups:
  • For the non-incentive group, the sample proportion \( \hat{p}_1 \) was calculated as \( \frac{75}{110} \approx 0.6818 \).
  • For the incentive group, the sample proportion \( \hat{p}_2 \) was \( \frac{66}{98} \approx 0.6735 \).
These sample proportions are used to compute the pooled proportion needed to assess the standard error and test statistics. Understanding sample proportions helps in determining how representative the sample is of the population and the reliability of the test results. It plays a pivotal role in the accuracy and interpretability of the statistical conclusions drawn from the hypothesis test.

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Most popular questions from this chapter

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