These Are the answer choices: A.) 343 B.) 342 C.) 432 D.) 677

As the sample size is large, i.e. n > 30, according to the Central limit theorem the sampling distribution of sample proportion will be normally distributed with mean $\hat p$ and standard deviation $\sqrt{\frac{\hat p (1-\hat p)}{n} }$ .

The mean is: $\hat p=0.31$

The confidence level (CL) = 98%

The confidence interval for single proportion is:

$CI_{p}=[\hat p-z_{(\alpha /2)}\times\sqrt{\frac{\hat p (1-\hat p)}{n} },\ \hat p+z_{(\alpha /2)}\times\sqrt{\frac{\hat p (1-\hat p)}{n} }]$

Here $z_{(\alpha /2)}$ = critical value and α = significance level.

Compute the value of α as follows:

$\alpha =1-CL\\=1-0.98\\=0.02$

For α = 0.02 the critical value can be computed from the z table.

Then the value of $z_{(\alpha /2)}$ is ± 2.33.

The 98% confidence interval for population proportion is:

$CI_{p}=[\hat p-z_{(\alpha /2)}\times\sqrt{\frac{\hat p (1-\hat p)}{n} },\ \hat p+z_{(\alpha /2)}\times\sqrt{\frac{\hat p (1-\hat p)}{n} }]\\=[0.31-2.33\times \sqrt{\frac{0.31\times(1-0.33)}{5046} },\ 0.31+2.33\times \sqrt{\frac{0.31\times(1-0.33)}{5046} } ]\\=[0.31-0.0152,\ 0.31+0.0152]\\=[0.2948,0.3252]\\\approx[0.30,\ 0.33]$

Thus, the 98% confidence interval [0.30, 0.33] implies that there is a 0.98 probability that the population proportion of people who refuse evacuation is between 0.30 and 0.33.

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3 years ago

These Are the answer choices:A.) 343 B.) 342 C.) 432D.) 677​

These Are the answer choices:A.) 343 B.) 342 C.) 432D.) 677