91Ó°ÊÓ

Two researchers conduct separate studies to test \(\mathrm{H}_{0}: p=0.50\) against \(\mathrm{H}_{a}: p \neq 0.50,\) each with \(n=400\) a. Researcher A gets 220 observations in the category of interest, and \(\hat{p}=220 / 400=0.550\) and test statistic \(z=2.00 .\) Show that the P-value \(=0.046\) for Researcher A's analysis. b. Researcher \(\mathrm{B}\) gets 219 in the category of interest, and \(\hat{p}=219 / 400=0.5475\) and test statistic \(z=1.90 .\) Show that the P-value \(=0.057\) for Researcher B's analysis. c. Using \(\alpha=0.05,\) indicate in each case from part a and part b whether the result is "statistically significant." Interpret. d. From part a, part b, and part c, explain why important information is lost by reporting the result of a test as "P-value \(\leq 0.05\) " versus "P-value \(>0.05\)," or as "reject \(\mathrm{H}_{0}\) " versus "do not reject \(\mathrm{H}_{0}\)," instead of reporting the actual P-value. e. Show that the \(95 \%\) confidence interval for \(p\) is (0.501,0.599) for Researcher \(\mathrm{A}\) and (0.499,0.596) for Researcher B. Explain how this method shows that, in practical terms, the two studies had very similar results.

Step by step solution

01

Calculate the Test Statistic for Researcher A

Researcher A has a sample proportion \( \hat{p} = 0.550 \). The null hypothesis proportion \( p_0 = 0.50 \). Using the formula for the z-test statistic: \[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} \] Plugging in the values: \[ z = \frac{0.550 - 0.50}{\sqrt{\frac{0.50 \cdot 0.50}{400}}} = 2.00 \] The test statistic for Researcher A is 2.00.

02

Determine the P-value for Researcher A

The P-value is the probability of observing a test statistic as extreme as, or more extreme than, the one observed, under the null hypothesis. For a two-tailed test with \( z = 2.00 \), we look up the standard normal distribution to find: \[ P(Z > 2.00) = 0.0228 \] Since it is two-tailed, the P-value is twice this: \[ \text{P-value} = 2 \times 0.0228 = 0.0456 \] Rounding to three decimal places, the P-value for Researcher A is 0.046.

03

Calculate the Test Statistic for Researcher B

Researcher B has a sample proportion \( \hat{p} = 0.5475 \) with the same null hypothesis proportion \( p_0 = 0.50 \). Using the z-test statistic formula: \[ z = \frac{0.5475 - 0.50}{\sqrt{\frac{0.50 \cdot 0.50}{400}}} = 1.90 \] Hence, the test statistic for Researcher B is 1.90.

04

Determine the P-value for Researcher B

For \( z = 1.90 \), we use the standard normal distribution to get the one side probability: \[ P(Z > 1.90) = 0.0287 \] The two-tailed P-value is: \[ \text{P-value} = 2 \times 0.0287 = 0.0574 \] Rounding to three decimal places, the P-value for Researcher B is 0.057.

05

Assess Statistical Significance

Using \( \alpha = 0.05 \): - For Researcher A, \( \text{P-value} = 0.046 \), which is less than 0.05, hence statistically significant (reject \( H_0 \)).- For Researcher B, \( \text{P-value} = 0.057 \), which is greater than 0.05, hence not statistically significant (do not reject \( H_0 \)).

06

Discuss the Importance of Actual P-values

By reporting only whether \( \text{P-value} \leq 0.05 \) or not, we lose precision about how strong the evidence is against \( H_0 \). Researcher A's result is at the boundary of significance, while Researcher B's result is just above it, indicating they arrived at similar conclusions but reporting only significance masks this.

07

Construct Confidence Intervals for Researcher A and B

To construct a 95% confidence interval for \( p \): \[ \hat{p} \pm z^* \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \]Where \( z^* = 1.96 \) for 95% CI:- For Researcher A: \[ 0.550 \pm 1.96 \times \sqrt{\frac{0.550 \times 0.450}{400}} = (0.501, 0.599) \]- For Researcher B: \[ 0.5475 \pm 1.96 \times \sqrt{\frac{0.5475 \times 0.4525}{400}} = (0.499, 0.596) \]The intervals show substantial overlap, demonstrating very similar results.

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Key Concepts

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

Confidence Intervals

When you think about confidence intervals, imagine you're setting a range to capture the true proportion of your population in a study. This range is calculated so that you can be reasonably sure—95% confident in most cases—that the true proportion lies somewhere within this interval. In the case of our two researchers, both constructed 95% confidence intervals for the proportion. Here’s how it works:
The formula for a confidence interval is: \[ \hat{p} \pm z^* \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \]where:

• \( \hat{p} \) is the sample proportion.
• \( n \) is the sample size.
• \( z^* \) is the z-value from the standard normal distribution for a given confidence level (1.96 for 95% confidence).

For Researcher A, the sample proportion \( \hat{p} \) was 0.550, leading to a confidence interval of (0.501, 0.599). For Researcher B, \( \hat{p} \) was 0.5475, with an interval (0.499, 0.596). Although their intervals are slightly different, notice how they overlap significantly. This indicates that, despite the small differences in their point estimates, the results of these two studies are exceptionally similar in practical terms.

P-value Interpretation

Understanding the P-value is crucial in hypothesis testing. Essentially, the P-value tells you how likely it is to observe your data, or something more extreme, given that the null hypothesis is true. In simpler terms, it's a measure that helps you judge whether there's enough evidence to reject the null hypothesis.
In our exercise, both researchers were testing whether the true proportion differed from 0.50. For Researcher A, the P-value was 0.046, and for Researcher B, it was 0.057. Here’s how to interpret these findings:

• If the P-value is less than or equal to 0.05, it's generally considered "statistically significant". This was the case for Researcher A, whose P-value was precisely at the standout of significance.
• For Researcher B, the P-value (0.057) was slightly above the 0.05 threshold, which means you would "not reject" the null hypothesis at this common level of significance.

While a small difference exists, knowing the exact P-values gives more insight into how strongly the data contradicts the null hypothesis. It highlights the precision lost when researchers only report results dichotomously as "significant" or "not significant".

Z-test for Proportions

The z-test for proportions is a statistical method used to determine whether there is a significant difference between a sample proportion and a hypothesized population proportion. This type of test is particularly handy when you have large sample sizes and you are interested in comparing proportions, as in the case of our researchers.
The standard formula for the z-test statistic is:\[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} \]where:

• \( \hat{p} \) is the sample proportion.
• \( p_0 \) is the hypothesized population proportion under the null hypothesis (in this case, 0.50).
• \( n \) is the sample size.

To calculate it, you substitute these values and simplify to find the z-value, which then helps you determine the P-value.

• For Researcher A, \( \hat{p} = 0.550 \) led to a z-value of 2.00.
• For Researcher B, \( \hat{p} = 0.5475 \) resulted in a z-value of 1.90.

These z-values tell you how many standard deviations away your sample proportion is from the hypothesized population proportion. It’s a critical step in assessing the strength of evidence against the null hypothesis, showcasing the power of the z-test in verifying claims about population parameters.