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Chapter 11: Problem 18

On May \(10-14,2001,\) the Gallup Organization surveyed 1002 adult Americans and asked them if they believed in psychic or spiritual healing. Of the 1002 individuals surveyed, 551 said yes. When the same question was asked on June \(6-8,2005,541\) of the 1002 individuals surveyed responded yes. (a) Test whether the proportion of adult Americans who believe in psychic or spiritual healing has changed since May 2001 at the \(\alpha=0.05\) level of significance. (b) Construct a \(90 \%\) confidence interval for the difference between the two population proportions, \(p_{2001}-p_{2005}\)

Short Answer

Expert verified
No significant change since May 2001. The 90% confidence interval for the difference between the two proportions is \((-0.0267, 0.0467)\).

Step by step solution

01

- Define Null and Alternative Hypotheses

Let \(p_{2001}\) be the proportion of adult Americans who believed in psychic or spiritual healing in 2001, and \(p_{2005}\) be the proportion in 2005. Null Hypothesis (\(H_0\)): \(p_{2001} = p_{2005}\) Alternative Hypothesis (\(H_1\)): \(p_{2001} eq p_{2005}\)
02

- Calculate Sample Proportions

Calculate the sample proportions for 2001 and 2005. \(\text{Sample proportion for 2001} (\hat{p}_{2001}) = \frac{551}{1002} \approx 0.5509\)\(\text{Sample proportion for 2005} (\hat{p}_{2005}) = \frac{541}{1002} \approx 0.5409\)
03

- Combine and Calculate the Pooled Proportion

Calculate the pooled proportion \(\hat{p}\) using the formula: \( \hat{p} = \frac{x_1 + x_2}{n_1 + n_2} = \frac{551 + 541}{1002 + 1002} = \frac{1092}{2004} \approx 0.545\)
04

- Calculate Standard Error

Calculate the standard error (SE) for the difference in proportions: \( SE = \sqrt{\hat{p}(1 - \hat{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)} = \sqrt{0.545(1 - 0.545) \left( \frac{1}{1002} + \frac{1}{1002} \right)} \approx 0.0223 \)
05

- Calculate the Test Statistic

Calculate the Z-test statistic using the formula: \( Z = \frac{\hat{p}_{2001} - \hat{p}_{2005}}{SE} = \frac{0.5509 - 0.5409}{0.0223} \approx 0.448\)
06

- Determine the Critical Value at \(\alpha = 0.05\)

The critical value for a two-tailed test at the 0.05 level of significance is approximately 1.96.
07

- Compare Test Statistic to Critical Value

Since \(|0.448| < 1.96\), fail to reject the null hypothesis. There is insufficient evidence to conclude that the proportion of adult Americans who believe in psychic or spiritual healing has changed since May 2001.
08

- Calculate Difference of Proportions for Confidence Interval

Calculate the difference between the sample proportions: \( \hat{p}_{2001} - \hat{p}_{2005} = 0.5509 - 0.5409 = 0.01 \)
09

- Calculate Margin of Error for 90% Confidence Interval

The Z value for a 90% confidence interval is approximately 1.645. Calculate the margin of error (ME): \( ME = Z_{0.05} \times SE = 1.645 \times 0.0223 \approx 0.0367 \)
10

- Construct 90% Confidence Interval

The confidence interval for the difference between the two population proportions is: \( 0.01 \pm 0.0367 \) which is \( (-0.0267, 0.0467)\)

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Key Concepts

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

Proportion Hypothesis Testing
Proportion hypothesis testing is a method used to compare a sample proportion to a known proportion, or to compare two sample proportions. Hypothesis testing helps to determine if the observed differences are statistically significant or if they occurred just by chance. Here, we used sample proportions of Americans who believed in psychic or spiritual healing in 2001 and 2005 to test if there was a significant change in belief over time.
Confidence Interval
A confidence interval is a range that estimates an unknown population parameter. For our exercise, we constructed a 90% confidence interval to estimate the difference between the proportions of 2001 and 2005. This interval, which calculated to be (-0.0267, 0.0467), tells us that we are 90% confident the true difference in proportions lies within this range. This helps us understand the precision and reliability of our estimate.
Z-test
The Z-test is a statistical method to determine if there is a significant difference between proportions. In our context, we used the Z-test to check if the proportion of Americans who believed in psychic or spiritual healing changed from 2001 to 2005. By calculating the Z-score (approximately 0.448), we compared it to the critical value (1.96 for a 0.05 significance level). Since our Z-score was less than the critical value, we concluded there was insufficient evidence to state that the proportions had changed significantly.
Standard Error
Standard error (SE) measures how much the sample proportion is expected to vary from the true population proportion. In our example, the standard error for the difference between the two proportions was calculated to be approximately 0.0223. This was an essential step as it helped us measure the variability of our sample and was used in both our hypothesis test and the construction of our confidence interval.

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Most popular questions from this chapter

Determine whether the sampling is dependent or independent. A study was conducted by researchers designed "to determine the genetic and nongenetic factors to structural brain abnormalities on schizophrenia."The researchers examined the brains of 29 patients diagnosed with schizophrenia and compared them with 29 healthy patients. The whole-brain volumes of the two groups were compared. (Source: William F. C. Baare et al., Volumes of Brain Structures in Twins Discordant for Schizophrenia, Archives of General Psychiatry 58: (2000) 33-40.)

Visual versus Textual Learners Researchers wanted to know whether there was a difference in comprehension among students learning a computer program based on the style of the text. They randomly divided 36 students into two groups of 18 each. The researchers verified that the 36 students were similar in terms of educational level, age, and so on. Group 1 individuals learned the software using a visual manual (multimodal instruction), while Group 2 individuals learned the software using a textual manual (unimodal instruction). The following data represent scores the students received on an exam given to them after they studied from the manuals. (a) What type of experimental design is this? (b) What are the treatments? (c) A normal probability plot and boxplot indicate it is reasonable to use Welch's \(t\) -test. Is there a difference in test scores at the \(\alpha=0.05\) level of significance? (d) Construct a \(95 \%\) confidence interval about \(\mu_{\text {visual }}-\mu_{\text {textual }}\) and interpret the results.

On April \(12,1955,\) Dr. Jonas Salk released the results of clinical trials for his vaccine to prevent polio. In these clinical trials, 400,000 children were randomly divided in two groups. The subjects in Group 1 (the experimental group) were given the vaccine, while the subjects in Group 2 (the control group) were given a placebo. Of the 200,000 children in the experimental group, 33 developed polio. Of the 200,000 children in the control group, 115 developed polio. (a) Test whether the proportion of subjects in the experimental group who contracted polio is less than the proportion of subjects in the control group who contracted polio at the \(\alpha=0.01\) level of significance. (b) Construct a \(90 \%\) confidence interval for the difference between the two population proportions, \(p_{1}-p_{2}\)

Determine whether the sampling is dependent or independent. An agricultural researcher wanted to determine whether there were any significant differences in the plowing method used on crop yield. He divided a parcel of land that had uniform soil quality into 30 subplots. He then randomIy selected 15 of the plots to be chisel plowed and 15 plots to be fall plowed. He recorded the crop yield at the end of the growing season to determine whether there was a significant difference in the mean crop yield.

Construct a confidence interval for \(p_{1}-p_{2}\) at the given level of confidence. \(x_{1}=804, n_{1}=874, x_{2}=892, n_{2}=954,95 \%\) confidence
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