Question

It is said that happy and healthy workers are efficient and productive. A company that manufactures exercising machines wanted to know the percentage of large companies that provide on-site health club facilities. A sample of such companies showed that of them provide such facilities on site. Construct a confidence interval for the percentage of all such companies that provide such facilities on site. What is the margin of error for this estimate? Round your answers to one decimal place.

Answer 1

Question;

It is said that happy and healthy workers are efficient and productive. A company that manufactures exercising machines wanted to know the percentage of large companies that provide on-site health club facilities. A sample of 200 such companies showed that 150 of them provide such facilities on site.

What is the point estimate of the percentage of all such companies that provide such facilities on site and Construct a 98% confidence interval for the percentage of all such companies that provide such facilities on site as well as find the margin of error for this estimate?

Answer:

Confidence interval is 0.7 ≤ p ≤ 0.8

The margin of error is 7.1 %

Step-by-step explanation:

Here we have, the proportion or point estimate given by

$\hat p$ = 150/200 = 0.75

Sample size, n = 200

The formula for confidence interval, CI, given a proportion is;

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

At 98% z = ±2.326348

Plugging in the values of $\hat p$ , z and n we get;

CI = 0.6787704 ≤ p ≤ 0.8212296

To one decimal place, we have

CI = 0.7 ≤ p ≤ 0.8

The margin of error = $z\times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$ , plugging in the values of $\hat p$ , z and n we get;

The margin of error = 0.0712296 = 7.12296%

The margin of error = 7.1 % to one decimal place.

Answer 2

Answer:

The 98% confidence interval for the population proportion is (0.7, 0.8).

The margin of error is MOE=0.071≈0.1.

Step-by-step explanation:

The question is incomplete. This is te complete question:

It is said that happy and healthy workers are efficient and productive. A company that manufactures exercising machines wanted to know the percentage of large companies that provide on-site health club facilities. A sample of 200 such companies showed that 150 of them provide such facilities on site.

Construct a 98% confidence interval for the percentage of all such companies that provide such facilities on site. What is the margin of error for this estimate? Round your answers to one decimal place.

We have to calculate a 98% confidence interval for the proportion.

The sample proportion is p=0.75.

$p=X/n=150/200=0.75$

The standard error of the proportion is:

$\sigma_p=\sqrt{\dfrac{\pi(1-\pi)}{n}}=\sqrt{\dfrac{0.75*0.25}{200}}\\\\\\ \sigma_p=\sqrt{0.001}=0.031$

The critical z-value for a 98% confidence interval is z=2.326.

The margin of error (MOE) can be calculated as:

$MOE=z\cdot \sigma_p=2.326 \cdot 0.031=0.071$

Then, the lower and upper bounds of the confidence interval are:

$LL=M-t \cdot s_M = 0.75-0.071=0.679\\\\UL=M+t \cdot s_M = 0.75+0.071=0.821$

The 98% confidence interval for the population proportion is (0.679, 0.821).