Question

The article "Breast-Feeding Rates Up Early" (USA Today, Sept. 14,2010 ) summarizes a survey of mothers whose babies were born in \(2009 .\) The Center for Disease Control sets goals for the proportion of mothers who will still be breast-feeding their babies at various ages. The goal for 12 months after birth is 0.25 or more. Suppose that the survey used a random sample of 1,200 mothers and that you want to use the survey data to decide if there is evidence that the goal is not being met. Let \(p\) denote the proportion of all mothers of babies born in 2009 who were still breast-feeding at 12 months. (Hint: See Example 10.10 ) a. Describe the shape, center, and spread of the sampling distribution of \(\hat{p}\) for random samples of size 1,200 if the null hypothesis \(H_{0}: p=0.25\) is true. b. Would you be surprised to observe a sample proportion as small as \(\hat{p}=0.24\) for a sample of size 1,200 if the null hypothesis \(H_{0}: p=0.25\) were true? Explain why or why not. c. Would you be surprised to observe a sample proportion as small as \(\hat{p}=0.20\) for a sample of size 1,200 if the null hypothesis \(H_{0}: p=0.25\) were true? Explain why or why not. d. The actual sample proportion observed in the study was \(\hat{p}=0.22 .\) Based on this sample proportion, is there convincing evidence that the goal is not being met, or is the observed sample proportion consistent with what you would expect to see when the null hypothesis is true? Support your answer with a probability calculation.

Short answer

The standard deviation is roughly 0.0128. A sample proportion of \(0.24\) is quite possible given the null hypothesis (\(p = 0.25\)), while \(0.20\) is rather surprising. The observed sample proportion of \(0.22\) provides some evidence against the null hypothesis, suggesting that it's not too surprising to have this sample proportion if the goal is met, but it's more likely to be observed when the actual population proportion is less than \(0.25\). This analysis does not conclusively prove that the goal is not being met, but it does provide an indication in that direction.

Step-by-step solution

Calculation of Standard Deviation

The shape of the sampling distribution will be approximately normal due to the large sample size (n = 1,200). The center of the distribution will be at \(p = 0.25\) (mean = \(p\)). The spread of the sampling distribution is characterized by its standard deviation, which we can calculate using the formula \(\sigma_{\hat{p}} = \sqrt{\frac{p(1 - p)}{n}}\), where \(p\) is the proportion under the null hypothesis and \(n = 1200\). The standard deviation, \(\sigma_{\hat{p}}\) is therefore, \(\sqrt{0.25(1 - 0.25)/1200} \approx 0.0128\).

Would you be surprised if \(\hat{p} = 0.24\)?

Next, we calculate the Z-score for the sample proportion \(0.24\) using the formula \(Z = \frac{\hat{p} - p}{\sigma_{\hat{p}}}\). Setting \(\hat{p} = 0.24\), we get \(Z = (0.24 - 0.25)/0.0128 = -0.78\). As the z-score is within -2 to 2, the sample proportion of \(0.24\) is not surprising given the null hypothesis of \(0.25\).

Would you be surprised if \(\hat{p} = 0.20\)?

We repeat the previous step with \(\hat{p} = 0.20\). The z-score we get is \(Z = (0.20 - 0.25)/0.0128 = -3.91\). This z-score lies outside the range of -2 to 2, therefore, we would be surprised to observe a sample proportion as small as \(0.20\). This suggests that if the true proportion were \(0.25\), it would be rare to observe a sample proportion of \(0.20\).

The sample proportion observed in the study

Finally, we need to calculate the Z-score for the sample proportion from the study, \(\hat{p} = 0.22\). This z-score is \(Z = (0.22 - 0.25)/0.0128 = -2.34\). Since this Z-score falls outside the range of -2 to 2 (though it's close), it provides some evidence against the null hypothesis. Although it's not too surprising to have this sample proportion if the goal is met, it's more likely to be observed when the actual population proportion is less than \(0.25\).