Question

construct a 90% confidence interval of the population proportion using the giver information x=74 n=150

Answer 1

The 90% confidence interval of the population proportion is (0.43, 0.56).

Explanation:

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

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

The information provided is:

X = 74

n = 150

Confidence level = 90%

Compute the value of sample proportion as follows:

`\hat p=(X)/(n)=(74)/(150)=0.493`

Compute the critical value of z for 90% confidence level as follows:

`z_(\alpha/2)=z_(0.10/2)=z_(0.05)=1.645`

*Use a z-table.

Compute the 90% confidence interval of the population proportion as follows:

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

`=0.493\pm 1.645* \sqrt{(0.493(1-0.493))/(150)}\\\\=0.493\pm 0.0672\\\\=(0.4258,\ 0.5602)\\\\\approx (0.43,\ 0.56)`

Thus, the 90% confidence interval of the population proportion is (0.43, 0.56).