Questions: A Canadian longitudinal study examined whether giving antibiotics in infancy increases the likelihood that the child will be overweight later in life. The study included 616 children and found that 438 of the children had received antibiotics during the first year of life. Test to see if this provides evidence that more than 70% of Canadian children receive antibiotics during the first year of life. Show all details of the hypothesis test, including hypotheses, the standardized test statistic, the p-value, the generic conclusion using a 5% significance level, and a conclusion in context. Clearly state the null and alternative hypotheses. Calculate the test statistic and p-value. Round your answer for the test statistic to two decimal places, and your answer for the p-value to three decimal places. What is the conclusion?

Solution Steps

We define the null and alternative hypotheses as follows:

• Null Hypothesis (\(H_0\)): \(p \leq 0.70\) (The proportion of Canadian children receiving antibiotics in the first year is 70% or less)
• Alternative Hypothesis (\(H_1\)): \(p > 0.70\) (The proportion of Canadian children receiving antibiotics in the first year is more than 70%)

The sample proportion (\(\hat{p}\)) is calculated as: \[ \hat{p} = \frac{x}{n} = \frac{438}{616} \approx 0.7110 \]

The test statistic (\(Z\)) is calculated using the formula: \[ Z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}} \] Substituting the values: \[ Z = \frac{0.7110 - 0.70}{\sqrt{\frac{0.70(1 - 0.70)}{616}}} \approx 0.5979 \]

The p-value associated with the test statistic is found to be: \[ \text{P-value} \approx 0.275 \]

At a significance level of \(\alpha = 0.05\), we compare the p-value to \(\alpha\):

• Since \(0.275 > 0.05\), we fail to reject the null hypothesis.

Thus, there is not enough evidence to conclude that more than 70% of Canadian children receive antibiotics during the first year of life.

Final Answer