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. check the success/failure condition. select the correct choice below and, if necessary, fill in the answer box to complete your choice. a. this condition is satisfied because ( n = square < 10% ) of the county population. b. this condition is satisfied because ( np = square geq 10 ) and ( nq = square geq 10 ). c. this condition is satisfied because the sample was a simple random sample. d. this condition is satisfied because ( n = square > 10% ) of the county population. e. this condition is satisfied because ( np = square leq 10 ) and ( nq = square leq 10 ). f. this condition is not satisfied.

Explanation:

Step1: Define hypotheses and parameters

Let $p$ = population proportion of obese adults.
$H_0: p = 0.358$, $H_a: p
eq 0.358$
Sample proportion: $\hat{p} = \frac{128}{400} = 0.32$

Step2: Check 10% Condition

The population is "a certain large county", so 400 is <10% of the population. This condition is satisfied.

Step3: Check Success/Failure Condition

Calculate $np$ and $nq$:
$np = 400 \times 0.358 = 143.2$
$nq = 400 \times (1-0.358) = 400 \times 0.642 = 256.8$
Both $\geq10$, so this condition is satisfied.

Step4: Calculate test statistic

$$z = \frac{\hat{p} - p}{\sqrt{\frac{p(1-p)}{n}}} = \frac{0.32 - 0.358}{\sqrt{\frac{0.358 \times 0.642}{400}}}$$
First compute denominator:
$\sqrt{\frac{0.358 \times 0.642}{400}} = \sqrt{\frac{0.229836}{400}} = \sqrt{0.00057459} \approx 0.02397$
Then $z = \frac{-0.038}{0.02397} \approx -1.585$

Step5: Find p-value

Two-tailed test, so p-value = $2 \times P(Z < -1.585) \approx 2 \times 0.0567 = 0.1134$

Step6: Compare p-value to significance level

The event is rare if p-value < 0.01. Here $0.1134 > 0.01$.

Step7: Correct the student's mistake

The student incorrectly assessed the 10% condition (400 <10% of a large county population, so the condition is satisfied, not the reason for failure). The correct Success/Failure condition is satisfied because $np=143.2 \geq10$ and $nq=256.8 \geq10$.

Answer:

Hypothesis Test Result:

No, the responses do not provide strong evidence that the 35.8% figure is inaccurate, as the p-value (0.1134) is greater than 0.01, so the event is not rare.

Corrected Success/Failure Condition:

B. This condition is satisfied because np = 143.2 ≥ 10 and nq = 256.8 ≥ 10.