Question

In a survey of 4060 ​adults, 717 oppose allowing transgender students to use the bathrooms of the opposite biological sex. Construct a​ 99% confidence interval for the population proportion. Interpret the results. A​ 99% confidence interval for the population proportion is ​( ____, ____​). ​(Round to three decimal places as​ needed.)

Part 2:

Use the given confidence interval to find the margin of error and the sample proportion. ​(0.631​, 0.659​)

Margin of Error=______ ​(Type an integer or a​ decimal.)

Sample Production=_____

Part 3:

Construct the indicated confidence interval for the population mean mu using the​ t-distribution. Assume the population is normally distributed.

c=0.99​,

x overbarx=14.8​,

s=2.0​,

n=8

(_____,____) ​(Round to one decimal place as​ needed.)

Answer 1

(1) = 717 / 4060 = 0.1766

The confidence interval = ME,

The Lower Limit = 0.1766 - 0.0154 = 0.1612    0.161

The Upper Limit = 0.1766 + 0.0154 = 0.192

The 99% CI is (0.161, 0.192)

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(2) - ME = 0.631

and + ME = 0.659

Adding we get, 2 * = 1.29

Therefore = Sample proportion = 0.645

Subtracting second equation from the first, we get

2 * ME = 0.028

ME = 0.014

Sample proportion = 0.645

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(3) The t critical value for df = n - 1 = 7 is 3.5

The CI is given by ME

Where ME = t critical * s/ sqrt(n) = 3.5 * 2 / sqrt(8) = 2.47

The Lower Limit = 14.8 - 2.47 = 12.33    12.3

The Upper Limit = 14.8 + 2.47 = 17.27    17.3

The CI is (12.3, 17.3)

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