Question

Find a 95% confidence interval for the true population proportion.

In a survey of 765 employees of a large company, 30% said that they had high job satisfaction.

Answer 1

To find the 95% confidence interval for the true population proportion, use the formula CI = p ± Z * √((p*q)/n), where p is the sample proportion, Z is the z-score, q is 1 - p, and n is the sample size. In this case, the 95% confidence interval is (0.26, 0.34).

Step-by-step explanation:

To find the 95% confidence interval for the true population proportion, we can use the formula:

CI = p ± Z * √((p*q)/n)

where:

• CI is the confidence interval
• p is the sample proportion
• Z is the z-score corresponding to the desired confidence level
• q is 1 - p
• n is the sample size

In this case, the sample proportion is 30%, the sample size is 765, and the z-score for a 95% confidence level is approximately 1.96.

Plugging these values into the formula, we get:

CI = 0.30 ± 1.96 * √((0.30 * 0.70)/765)

Simplifying this expression gives:

CI = (0.26, 0.34)

Therefore, the 95% confidence interval for the true population proportion is (0.26, 0.34).

Answer 2

The 95% confidence interval for the true population proportion is (0.27, 0.33).

Step-by-step explanation:

The (1 - α)% confidence interval for population proportion of is:

`CI=\hat p \pm z_(\alpha/2)\cdot\sqrt{(\hat p(1-\hat p))/(n)}`

The information given is:

`\hat p=0.30\\n=765\\\text{Confidence level}=95\%`

The critical value of z for 95% confidence level is, 1.96.

Compute the 95% confidence interval for the true population proportion as follows:

`CI=\hat p \pm z_(\alpha/2)\cdot\sqrt{(\hat p(1-\hat p))/(n)}`

`=0.30\pm 1.96*\sqrt{(0.30(1-0.30))/(765)}\\\\=0.30\pm 0.0325\\\\=(0.2675, 0.3325)\\\\\approx (0.27, 0.33)`

Thus, the 95% confidence interval for the true population proportion is (0.27, 0.33).