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.

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 = 143 geq 10 ) and ( nq = 257 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.

the observed proportion of successes in the sample is ( \frac{square}{square} = square ).

(round to three decimal places as needed.)

Explanation:

Step1: Calculate observed sample proportion

$\hat{p} = \frac{128}{400} = 0.320$

Step2: Verify Success/Failure Condition

First, define $p=0.358$, $n=400$, $q=1-p=0.642$.
Calculate $np = 400 \times 0.358 = 143.2$, $nq = 400 \times 0.642 = 256.8$. Both are $\geq10$.

Step3: Verify 10% Condition

The sample size $n=400$ is less than 10% of the population (a "large country" has far more than 4000 adults).

Step4: Set up hypothesis test

We test $H_0: p=0.358$, $H_a: p
eq 0.358$.
Calculate standard error: $SE = \sqrt{\frac{pq}{n}} = \sqrt{\frac{0.358 \times 0.642}{400}} \approx 0.0239$
Calculate z-score: $z = \frac{\hat{p}-p}{SE} = \frac{0.320-0.358}{0.0239} \approx -1.59$

Step5: Find p-value

For two-tailed test, $p\text{-value} = 2 \times P(Z < -1.59) = 2 \times 0.0559 = 0.1118$
This p-value is greater than 0.01, so we fail to reject $H_0$.

Answer:

Observed Proportion:

$\frac{128}{400} = 0.320$

Success/Failure Condition:

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

Hypothesis Test Conclusion:

The p-value (0.112) is greater than 0.01, so there is not strong evidence that the 35.8% figure is inaccurate for this region.