Question

Of 118 randomly selected adults, 34 were found to have high blood pressure. Construct a 95% confidence interval for the true percentage of all adults that have high blood pressure. Construct a confidence interval for the population proportion p.

Answer 1

The confidence interval is
`0.20644 < p <0.36984`

Explanation:

From the question we are told that

The sample is n = 118

The confidence level is C = 95 %

The number of people with high blood pressure is k = 34

The proportion of those with high blood pressure is evaluated as

`\r p = (k)/(n)`

substituting values

`\r p = (34)/(118)`

`\r p = 0.288136`

Given that the confidence level is 95% then the level of significance is evaluated as

`\alpha = 100 -95`

`\alpha = 5`%

`\alpha = 0.05`

Now the critical values of
`(\alpha )/(2)` obtained from the normal distribution table is
`Z_{(\alpha )/(2) } = 1.96`

The reason we are obtaining values for is because is the area under the normal distribution curve for both the left and right tail where the 95% interval did not cover while is the area under the normal distribution curve for just one tail and we need the value for one tail in order to calculate the confidence interval

Now the margin of error is evaluated as

`MOE = Z_{(\alpha )/(2) } * \sqrt{(\r p (1- \r p))/(n) }`

substituting values

`MOE = 1.96 * \sqrt{( 0.288136 (1- 0.288136))/(118) }`

`MOE = 0.0817`

Thus the 95% confidence interval for the true percentage of all adults that have high blood pressure is evaluated as

`\r p - MOE < p < \r p + MOE`

substituting values

`0.288136 - 0.0817 < p <0.288136 + 0.0817`

`0.20644 < p <0.36984`