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 240 such companies showed that 96 of them provide such facilities. Construct a 97% confidence interval for the percentage of all such companies that provide such facilities on-site. What is the margin of error for this estimate

Answer 1

To construct a 97% confidence interval for the percentage of large companies that provide on-site health club facilities, the proportion p is calculated from the sample data and then substituted into the formula. The margin of error is also calculated using the formula.

Step-by-step explanation:

To construct a 97% confidence interval for the percentage of large companies that provide on-site health club facilities, we can use the formula:

CI = p ± z*(√(p(1-p)/n))

Where:

• p is the proportion of companies in the sample that provide on-site health club facilities
• z is the z-value corresponding to the desired confidence level (in this case, 97%)
• n is the sample size

First, we calculate the proportion p = 96/240 = 0.4

Next, we find the z-value using a standard normal distribution table. For a 97% confidence level, the z-value is approximately 1.88

Finally, substituting the values into the formula:

CI = 0.4 ± 1.88*(√((0.4*(1-0.4))/240))

Calculating the margin of error:

ME = z*(√(p(1-p)/n))

ME = 1.88*(√((0.4*(1-0.4))/240))

ME ≈ 0.036

Therefore, the 97% confidence interval for the percentage of all such companies that provide on-site health club facilities is approximately 0.364 to 0.436. The margin of error for this estimate is approximately ±0.036.

Answer 2

The 97% confidence interval for the percentage of all such companies that provide such facilities on-site is (0.3314, 0.4686). The margin of error is of 0.0686 = 6.86 percentage points.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of
`\pi`, and a confidence level of
`1-\alpha`, we have the following confidence interval of proportions.

`\pi \pm z\sqrt{(\pi(1-\pi))/(n)}`

In which

z is the zscore that has a pvalue of
`1 - (\alpha)/(2)`.

The margin of error is:

The absolute value of the subtraction of one of the bounds by the estimate
`\pi`

For this problem, we have that:

`n = 240, \pi = (96)/(240) = 0.4`

97% confidence level

So
`\alpha = 0.03`, z is the value of Z that has a pvalue of
`1 - (0.03)/(2) = 0.985`, so
`Z = 2.17`.

The lower limit of this interval is:

`\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.4 - 2.17\sqrt{(0.4*0.6)/(240)} = 0.3314`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.4 + 2.17\sqrt{(0.4*0.6)/(240)} = 0.4686`

0.4686 - 0.4 = 0.0686

The 97% confidence interval for the percentage of all such companies that provide such facilities on-site is (0.3314, 0.4686). The margin of error is of 0.0686 = 6.86 percentage points.