Question

in a recent year, a health agency reported that 35.8% of adults in a certain large country are obese. a county health service planning a new awareness campaign polls a random sample of 400 adults living there. in this sample, 128 people were found to be obese based on their answers to a health questionnaire. do these responses provide strong evidence that the 35.8% figure is not accurate for this region? correct the mistakes you find in the accompanying student’s attempt to test an appropriate hypothesis. consider an event to be rare if its probability of occurring is less than 0.01. show how the p - value will be found. select the correct choice below and fill in the answer box to complete your choice. (round to two decimal places as needed) a. ( p = p(z > ) () ) b. ( p = p(z < ) () ) c. ( p = 2p(z < - 1.59) ) d. ( p = 2p(t < ) () ) e. ( p = p(t < ) () ) f. ( p=\frac{1}{2}p(z < ) () ) g. ( p = p(t > ) () ) h. ( p=\frac{1}{2}p(t < ) () ) the p - value is (round to three decimal places as needed).

Explanation:

Step1: State hypotheses

$H_0: p = 0.358$, $H_a: p
eq 0.358$ (two-tailed test)

Step2: Calculate sample proportion

$\hat{p} = \frac{128}{400} = 0.32$

Step3: Compute test statistic

$$z = \frac{\hat{p}-p}{\sqrt{\frac{p(1-p)}{n}}} = \frac{0.32-0.358}{\sqrt{\frac{0.358(1-0.358)}{400}}} \approx -1.59$$

Step4: Identify correct P-value form

For two-tailed z-test: $P=2P(z < -1.59)$

Step5: Calculate P-value

$P(z < -1.59) = 0.0559$, so $P=2\times0.0559=0.1118$

Answer:

C. $P=2P(z < -1.59)$
The P-value is 0.112