Question

A random sample of residents in city J were surveyed about whether they supported raising taxes to increase bus service for the city. From the results, a 95 percent confidence interval was constructed to estimate the proportion of people in the city who support the increase. The interval was ____________

Answer 1

C. More than 40 percent of the residents support the increase.

Explanation:

quizlet/chegg

Answer 2

CI (at 95%) =
`p \pm Z_(0.05/2) *\sqrt{(p(1-p))/(n) }`

Explanation:

The question does not provide any numeric observation to compute the interval. Consequently, we have to do so symbolically or using the formula and describe the formula.

The formula is given as:

CI (at 95%) =
`p \pm Z_(0.05/2) *\sqrt{(p(1-p))/(n) }`

Where:

p is the proportion of of people in the city who support the increase.

(1-p) is the proportion of of people in the city who do not support the increase.

n is the sample size selected during the survey.

`Z_(\alpha/2)`, where
`\alpha` = 5%., the
`Z_(\alpha/2)` is the table value obtain from the normal distribution table. This is 1.96 in this case!

Therefore, if we supply each value and substitute them into the formula, we would obtain the interval.