Question

In a random sample of 316 people that were tested for the Norcovirus, it was found that 262 did not have the virus. Construct a 95% confidence interval to estimate the proportion of the population that does not have the Norcovirus. a. Which parameter are you estimating? р X b. What is the point estimate? (Round your answer to four decimal places) = 0.8291 P c. Check all of the requirements that are satisfied. random the 7 distribution is normal since n > 30 the i distribution is normal since the distribution is normal the p distribution is normal since no > 10 and nĝ > 10 d. Which calculator function should you use? 1-PropZTest х e. Find the margin of error. Round your answer to four decimal places. E 0.0431 f. Give the confidence interval in interval notation (0.1234, 0.4321). Use decimals, not percentages and round your answer to four decimal places. (0.7860,0.8722) g. Select the correct condusion. We are 95% confident that the proportion of people in the sample that tested postitive for the Norcovirus is between the two proportions found. We are 95% confident that the proportion of people in the sample that tested negative for the Norcovirus is between the two proportions found. We are 95% confident that the proportion of the population that have the Norcovirus is between the two proportions found. We are 95% confident that the proportion of the population that does not have the Norcovirus is between the two proportions found.

Answer 1

Answer:

(a)

The parameter we are estimating is

People who does not have the Norcovirus

(b)

Point of estimate :-

$n=316$

$X=262$

262 Р = 0.8291 n 316

(c)

Option D :
The distributed is normal since np > 10 and ng > 10

(d)

Z-table distribution function we will use

(e)

Margin of Error:-

$E=Z_{\frac{\alpha}{2}}\times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

Here:-

Confidence interval = 0.95

$\alpha=1-0.95=0.05$

$\frac{\alpha}{2}=0.025$

$Z_{critical}=Z_{\frac{\alpha}{2}}=Z_{0.025}=\pm 1.645\ (using\ Z-table)$

$E=1.96\times \sqrt{\frac{0.8291(1-0.8291)}{316}}$

$E=0.0415\approx 0.0431$

(f)

Confidence interval :-

$=\hat{p}\pm E(Margin\ of\ error)$

$=0.8291\pm 0.0415$

${\color{Green} \mathbf{=(0.7860,0.8722)}}$

(g)

Option (d) : We are 95% confident that the proportion of the population that does not have the Norcovirus is between the two proportions founds.