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Significant Does Not Mean Important. Never forget that even small effects can be statistically significant if the samples are large. To illustrate this fact, consider a sample of 148 small businesses. During a three-year period, 15 of the 106 businesses headed by men and seven of the 42 businesses headed by women failed. 22 a. Find the proportions of failures for businesses headed by women and businesses headed by men. These sample proportions are quite close to each other. Give the \(P\). value for the \(z\) test of the hypothesis that the same proportion of women's and men's businesses fail. (Use the two-sided alternative.) The test is very far from being significant. b. Now suppose that the same sample proportions came from a sample 30 times as large. That is, 210 out of 1260 businesses headed by women and 450 out of 3180 businesses headed by men fail. Verify that the proportions of failures are exactly the same as in part (a). Repeat the \(z\) test for the new data and show that it is now significant at the \(\alpha=0.05\) level. c. It is wise to use a confidence interval to estimate the size of an effect rather than just giving a P-value. Give the large sample \(95 \%\) confidence intervals for the difference between the proportions of women's and men's businesses that fail for the settings of both parts (a) and (b). What is the effect of larger samples on the confidence interval? Do you think the size of the difference between the proportions is an important difference? In responding to Exercises \(23.26\) through \(23.36\) follow the Plan, Solve, and Conclude steps of the four-step process.

Step by step solution

01

Calculate Failure Proportions (Part a)

For businesses headed by men: - Total businesses: 106- Failed businesses: 15 - Proportion: \( p_{men} = \frac{15}{106} \approx 0.1415 \)For businesses headed by women: - Total businesses: 42- Failed businesses: 7 - Proportion: \( p_{women} = \frac{7}{42} \approx 0.1667 \)

02

Perform z-Test (Part a)

Calculate pooled proportion \( p \) for the two groups:\[ p = \frac{15+7}{106+42} = \frac{22}{148} \approx 0.1486 \]Calculate \( z \) statistic:\[ z = \frac{0.1667 - 0.1415}{\sqrt{0.1486 \cdot (1-0.1486) \cdot \left(\frac{1}{106} + \frac{1}{42}\right)}} \approx 0.547 \]Using a standard normal distribution table, the two-tailed p-value corresponding to \( z = 0.547 \) is not significant.

03

Verify Proportions for Larger Sample (Part b)

Verify the same proportions for a larger sample:For women: - Total businesses: 1260- Failed businesses: 210- Proportion: \( \frac{210}{1260} = 0.1667 \)For men: - Total businesses: 3180- Failed businesses: 450- Proportion: \( \frac{450}{3180} = 0.1415 \)The proportions match those from part (a).

04

Perform z-Test for Larger Sample (Part b)

Pooled proportion for the larger sample:\[ p = \frac{450+210}{3180+1260} = \frac{660}{4440} \approx 0.1486 \]Calculate \( z \) statistic for larger sample:\[ z = \frac{0.1667 - 0.1415}{\sqrt{0.1486 \cdot (1-0.1486) \cdot \left(\frac{1}{3180} + \frac{1}{1260}\right)}} \approx 2.997 \]Using a standard normal distribution, the p-value is less than \( 0.05 \), indicating a significant difference.

05

Confidence Interval Calculation (Part c)

For part (a):Calculate standard error (SE):\[ SE_{a} = \sqrt{0.1486 \cdot (1-0.1486) \cdot \left(\frac{1}{106} + \frac{1}{42}\right)} \approx 0.0637 \]The confidence interval:\[ (0.1667 - 0.1415) \pm 1.96 \times 0.0637 \approx ( -0.067, 0.116) \]For part (b):Calculate standard error (SE):\[ SE_{b} = \sqrt{0.1486 \cdot (1-0.1486) \cdot \left(\frac{1}{3180} + \frac{1}{1260}\right)} \approx 0.0214 \]The confidence interval:\[ (0.1667 - 0.1415) \pm 1.96 \times 0.0214 \approx (0.004, 0.050) \]The larger sample size results in a narrower confidence interval.

06

Conclusion (Part c)

Larger sample sizes provide more precise estimates of effect size, as indicated by the narrower confidence interval in part (b). The actual difference in proportions remains small, suggesting that the difference may not be practically important.

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Key Concepts

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

z-test

In statistics, a z-test is a powerful tool used to determine if there are significant differences between sample and population means. It compares the observed data with what the null hypothesis suggests. In our scenario, we want to see if the proportion of business failures led by men differs significantly from those led by women.
A z-test involves several key steps:

• First, we calculate the sample proportions, which are the ratios of failed businesses to total businesses for each group.


• Next, we compute the pooled proportion of failures for both groups combined. This step helps in standardizing the differences.


• Then, we calculate the z-score using the difference between sample proportions and the standard error derived from the pooled proportion. The formula for the z-score is:\[z = \frac{\text{Observed Proportion Difference} - 0}{\text{Standard Error}}\]


• Finally, by comparing this z-score against a standard distribution table, we obtain a p-value, which indicates how significant the observed differences are.

In the exercise above, the z-test performed with the initial smaller sample size wasn't significant, but when the sample size increased dramatically, the z-test showed statistical significance.

confidence interval

Confidence intervals give us a range of values that can estimate the true population parameter. They are particularly useful when you want to understand how certain you can be about your sample results.
A 95% confidence interval tells us that, if we were to take 100 different samples and compute confidence intervals for each, about 95 of those intervals would contain the true population proportion.
To create a confidence interval around the difference in proportions of business failures:

• The standard error for the difference in proportions is computed using the pooled proportion from both groups.


• The margin of error is then estimated by multiplying this standard error with the z-score that corresponds to the desired confidence level, often 1.96 for 95% confidence.


• The confidence interval itself is calculated as the observed difference in sample proportions plus or minus this margin of error. \[CI = (\text{Proportion Difference}) \pm (z \times SE)\]

With a larger sample size, the confidence interval becomes narrower, indicating a more precise estimate. Despite the precision, the actual practical significance of the difference in this scenario remains a debate.

sample size effect

The size of your sample can greatly influence the outcome of statistical tests, such as the z-test. With bigger samples, we tend to see more reliable estimates and often higher statistical significance. This is due to the reduction of variability and the ability to detect even small differences.
In the given exercise, when we increased the sample size by 30 times, from 148 to 4440 businesses, the previously non-significant z-test became significant.
Key impacts of sample size include:

• **Precision:** Larger sample sizes lead to more precise estimates, which shrink the standard error and result in narrower confidence intervals. This precision can help determine effects more accurately.


• **Power:** The ability of a test to detect differences (its power) increases with larger sample sizes, meaning smaller differences might become statistically significant.


• **Significance Testing:** A difference might be statistically significant in a large sample, but it doesn't always mean the difference is practically significant.

While larger sample sizes lend greater statistical power, it's essential to interpret the significance in the context of practical application. Even if the statistics suggest significance, real-world implications must always be considered.