Question

in a recent year, a health agency reported that 35.8% of adults in a certain large county 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. click the icon to view the student’s attempt to test an appropriate hypothesis. \\( h_0: p = 0.358 \\) \\( h_a: p \
eq 0.358 \\) (type integers or decimals. do not round.) check the independence assumption and the randomization condition. select the correct choice below and, if necessary, fill in the answer box to complete your choice. a. these are satisfied because the sample was a simple random sample. b. these are satisfied because \\( n = \\) \\( < 10\\% \\) of the county population. c. these are satisfied because \\( np = \\) \\( \leq 10 \\) and \\( nq = \\) \\( \leq 10 \\). d. these are both egregiously violated. e. these are satisfied because \\( n = \\) \\( > 10\\% \\) of the county population. f. these are satisfied because \\( np = \\) \\( \geq 10 \\) and \\( nq = \\) \\( \geq 10 \\).

Explanation:

Step1: Calculate np

$np = 400 \times 0.358 = 143.2$

Step2: Calculate nq (q=1-p)

$nq = 400 \times (1-0.358) = 400 \times 0.642 = 256.8$

Step3: Check 10% condition

The population is a "certain large country", so 400 is <10% of the total adult population.

Step4: Evaluate assumptions

Randomization is satisfied (random sample). np and nq are both ≥10, and n <10% of population.

Answer:

E. These are satisfied because n = 400 < 10% of the county population.
F. These are satisfied because np = 143.2 ≥ 10 and nq = 256.8 ≥ 10.

---

Hypothesis Test Completion (full test for context):

Step5: Calculate sample proportion

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

Step6: Compute test statistic

$$z = \frac{\hat{p} - p}{\sqrt{\frac{p(1-p)}{n}}} = \frac{0.32 - 0.358}{\sqrt{\frac{0.358 \times 0.642}{400}}} = \frac{-0.038}{\sqrt{0.000574}} \approx -1.59$$

Step7: Find p-value (two-tailed)

$p\text{-value} = 2 \times P(Z < -1.59) = 2 \times 0.0559 = 0.1118$

Step8: Compare to significance level

$0.1118 > 0.01$, so fail to reject $H_0$.

Final Hypothesis Test Answer:

There is not strong evidence at the 0.01 significance level that the 35.8% figure is inaccurate for the region.