10.3 - Tests for Differences

• Null Hypothesis: The gender of the evaluator does not affect the population proportion of evaluators who prefer the female candidate \(( p_{\text{females}} = p_{\text{males}})\) or \(( p_{\text{females}} - p_{\text{males}}=0)\)
• Alternative Hypothesis: The gender of the evaluator does affect the population proportion of evaluators who prefer the female candidate \(( p_{\text{females}} \neq p_{\text{males}})\) or \(( p_{\text{females}} - p_{\text{males}}\neq 0)\) This is a two sided alternative.

The sample proportion for the female evaluators was 0.703 while the sample proportion for the male evaluators was 0.425; a difference of 0.703 - 0.425 = 0.278. If the null hypothesis is true then p_females = p_males and the best estimate of this common overall probability of preferring the divorced female candidate would be 0.549. Thus, the standard error for the difference in proportions under the null hypothesis would be

Thus the standardized test statistic would be (0.278 - 0) / 0.0834 = 3.33.

Since the sample size is fairly large for each group the difference between the two sample proportions would follow the normal curve. Since this is a two-sided alternative, we calculate the p-value by considering both the area above 3.33 and the area below -3.33 on the normal curve (this comes out to about 0.00045 + 0.00045 = 0.0009).

Interpretation of the p-value. The likelihood of getting our test statistic of 3.33 or any more extreme value (like those above it or below -3.33), if in fact, the null hypothesis is true, is about 0.0009; a bit less than one-tenth of one percent.

Since the p-value is so small the results are highly significant; the null hypothesis provides a poor explanation of the data. We have good evidence that there is an association between the gender of the evaluator and the preferences they would hold between candidates with these gender and lifestyle combinations.

• Null Hypothesis: The color of the mug does not affect the population average aroma rating (mean_blue = mean_{white or} mean_blue - mean_{blue = 0}).
• Alternative Hypothesis: The color of the mug does affect the population average aroma rating (mean_blue = mean_{white or} mean_blue - mean_{blue = 0}). This is a two-sided alternative.

The sample mean for the white cups was 57.33 while the sample mean for the blue cups was 35.57; a difference of 57.33 - 35.57 = 21.76. The standard error of the mean for the white cups was \(\frac{16.27}{\sqrt{12}}= 4.70\) while the standard error for the blue cups was \(\frac{25.34}{\sqrt{12}}= 7.32\). Thus, the standard error for the difference in proportions under the null hypothesis would be \(\sqrt{(4.70)^{2} + (7.32)^{2}}= 8.70\).

Finally, the standardized test statistic would be (21.76 - 0) / 8.70 = 2.5

While the difference in the sample means might nearly follow the normal curve, the estimate of the standard error of the differences might be a bit off from the actual standard error so the use of the t-curve, rather than the normal curve, would be the appropriate reference distribution to calculate the p-value. Since this is a two-sided alternative, we calculate the p-value by considering both the area above 2.5 and the area below -2.5 on the t curve with sample sizes of 12 in each group (this comes out to about 0.01 + 0.01 = 0.02).

Interpretation of the p-value. The likelihood of getting our test statistic of 2.5 or any more extreme value (like those above it or below -2.5), if, in fact, the null hypothesis is true, is about 2%.

Since the p-value is less than 5% so the results might be considered significant; the null hypothesis provides a fairly poor explanation of the data. We have some evidence that there is an association between the color of the mug and the perceived aroma of the coffee.