Problem 32 - Basic Statistics for Business and Economics with Student

Question

A study was conducted to determine if there was a difference in the humor content in British and American trade magazine advertisements. In an independent random sample of 270 American trade magazine advertisements, 56 were humorous. An independent random sample of 203 British trade magazines contained 52 humorous ads. Does this data provide evidence at the .05 significance level that there is a difference in the proportion of humorous ads in British versus American trade magazines?

Step-by-Step Solution

Answer

No, there isn't enough evidence to suggest a difference.

Step 1: Define Hypotheses

First, we establish the null and alternative hypotheses. The null hypothesis (\( H_0 \) ) is that there is no difference in the proportion of humorous ads between British and American magazines, i.e.,\( p_1 = p_2 \). The alternative hypothesis (\( H_a \) ) is that there is a difference, i.e.,\( p_1 eq p_2 \).

Step 2: Calculate Sample Proportions

Calculate the sample proportions of humorous ads for British and American magazines. For British magazines, the sample proportion is\( \hat{p}_1 = \frac{52}{203} \approx 0.2563 \). For American magazines, the sample proportion is\( \hat{p}_2 = \frac{56}{270} \approx 0.2074 \).

Step 3: Calculate Pooled Proportion

Since we are comparing two proportions, we calculate the pooled proportion. The pooled proportion is given by:\[\hat{p} = \frac{x_1 + x_2}{n_1 + n_2} = \frac{52 + 56}{203 + 270} = \frac{108}{473} \approx 0.2284\]

Step 4: Calculate Standard Error

Next, calculate the standard error (SE) for the difference in proportions using the pooled proportion:\[SE = \sqrt{\hat{p}(1-\hat{p})\left(\frac{1}{n_1} + \frac{1}{n_2}\right)} = \sqrt{0.2284 \times 0.7716 \times \left(\frac{1}{203} + \frac{1}{270}\right)} \approx 0.0458\]

Step 5: Calculate Z-Score

Using the sample proportions and the standard error, compute the Z-score:\[Z = \frac{\hat{p}_1 - \hat{p}_2}{SE} = \frac{0.2563 - 0.2074}{0.0458} \approx 1.067\]

Step 6: Determine P-Value and Conclusion

Check the Z-score against the standard normal distribution. A two-tailed test at a significance level of 0.05 corresponds to a critical value of approximately 1.96. Since the calculated Z-score (1.067) is less than 1.96, we fail to reject the null hypothesis. Therefore, there is not enough evidence to suggest a significant difference in the proportion of humorous ads between British and American trade magazines at the 0.05 significance level.

Key Concepts

Statistical Significance

Statistical significance helps us determine if our findings from a sample can be extended to the larger population. In simple terms, it tells us whether the observed difference or relationship in our data is genuine or if it could have happened by chance. In the case of the humorous ads study, we want to know if the difference in proportions of humorous ads between American and British magazines is likely due to true differences or random variability.
For our study, we chose a significance level of 0.05. This is a common benchmark in statistical testing, allowing a 5% chance that our conclusion might be incorrect. If our calculated value (Z-score) lies beyond the critical values for this level, we say the result is statistically significant. Otherwise, we conclude that the effect is insignificant, indicating no strong evidence against the null hypothesis.

Sample Proportions

Sample proportions are estimates that help us understand how often a particular characteristic appears within a sample group. In this problem, we're interested in the proportion of humorous ads among the sampled magazines from Britain and America.

• For British magazines: The sample proportion is given by the expression \( \hat{p}_1 = \frac{52}{203} \approx 0.2563 \).
• For American magazines: The sample proportion is \( \hat{p}_2 = \frac{56}{270} \approx 0.2074 \).

These proportions tell us that out of the sampled ads, approximately 25.63% of British ads and 20.74% of American ads were humorous. Sample proportions are crucial when performing hypothesis testing, as they represent the fundamental data used to determine if our observations of difference are statistically significant.

Pooled Proportion

When comparing two sample proportions, it's often useful to consider a pooled proportion. This combined proportion gives a single measure of the characteristic of interest based on the total samples from both groups.In our problem, the pooled proportion is:\[ \hat{p} = \frac{x_1 + x_2}{n_1 + n_2} = \frac{52 + 56}{203 + 270} = \frac{108}{473} \approx 0.2284 \]This value represents the overall proportion of humorous ads across both British and American magazines in the sampled data. By using this single value, we can more accurately estimate the variability when calculating the standard error, which in turn, helps in determining the Z-score for testing the hypothesis.

Z-Score

The Z-score is a statistical metric that tells us how many standard deviations an element is from the mean of the dataset. In hypothesis testing, it’s a crucial step in determining whether there's evidence to reject the null hypothesis.To compute the Z-score in this context, we use the formula:\[ Z = \frac{\hat{p}_1 - \hat{p}_2}{SE} = \frac{0.2563 - 0.2074}{0.0458} \approx 1.067 \]Here, \( \hat{p}_1 \) and \( \hat{p}_2 \) are the sample proportions, and \( SE \) is the standard error calculated using the pooled proportion.A Z-score of 1.067 indicates that the difference in proportions is only about 1.067 standard deviations away from what we might expect under the null hypothesis. When we compare this Z-score to the critical value for a 0.05 significance level (which is 1.96 for a two-tailed test), we find it is lower, suggesting insufficient evidence to reject the null hypothesis. Hence, the data does not provide significant evidence of a difference in the humor content of magazine ads between the two countries.