Question

The workers at XYZ, Inc. took a random sample of 800 manhole covers and found that A) (37.26, 42.74) 40 of them were defective. What is the 95 percent Conf. Interval for p, the true proportion of defective manhole covers, based on this sample? SELECT ALL APPLICABLE CHOICES B) (.015,.085) D) (047, 053) C) (035,.065)

Answer 1

Ans. The correct option is (C) 0.035,0.065

First in order to calculate the confidence interval for the true proportion of defective manhole covers from the sample of 800 manhole covers out of which 40 are defective. We have to first calculate the sample proportion i.e., $\hat{p}$.

The formula to calculate the $\hat{p}$ is :   $\hat{p}=\frac{x}{n}$

In this case

x: number of defective manhole cover.

n: total number of manhole cover.

x=40, n=800

$\hat{p}=\frac{x}{n}=\frac{40}{800}=0.05$

$\mathbf{\hat{p}=0.05}$

Now the formula to calculate the confidence interval for population/true proportion is:

$\hat{p}\pm z\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

we have calculated the sample proportion $\hat{p}=0.05$, now we have to just plugin the values, and the value of z for 95% confidence interval is 1.96.

$\hat{p}\pm z\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

$0.05\pm 1.96\sqrt{\frac{0.05(1-0.05)}{800}}$

$0.05\pm 1.96\sqrt{\frac{0.05(0.95)}{800}}$

$0.05\pm 1.96\sqrt{\frac{0.0475}{800}}$

$0.05\pm 1.96\sqrt{0.000059375}$

$0.05\pm 1.96(0.0077)$

$0.05\pm 0.0151$

$(0.05-0.0151, 0.05+0.0151) or \mathbf{(0.035, 0.065)}$

The 95% confidence interval for the true proportion of defective manhole cover is $\mathbf{0.035, 0.065}$ , or we can say that we are 95% confident that the true proportion of defective manhole cover is between 0.035 and 0.065.

Ans. The correct option is   B) The expected value

For any given variable x the mathematical expectation of x is defined as "the sum of products of the values of x with the corresponding probabilities."

Algebraic form is:    $E(x)=\sum_{i=1}^{n}x_{i}p_{i}=x_{1}p_{1}+x_{2}p_{2}+........+x_{n}p_{n}$