Assume a 2015 Gallup Poll asked a national random sample of 515 adult women to state...

Assume a 2015 Gallup Poll asked a national random sample of 515 adult women to state their current weight. Assume the mean weight in the sample was ?¯=155.

We will treat these data as an SRS from a normally distributed population with standard deviation ?=33 pounds.

Give a 99% confidence interval for the mean weight of adult women based on these data. Enter the upper and lower values of your confidence interval into the spaces provided rounded to two decimal places.

Solutions

Expert Solution

solution :

Given that,

= 155

= 33

n = 515

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576

Margin of error = E = Z_/2* ( / $\sqrt$ n)

= 2.576* ( 33/ $\sqrt$ 515)

= 3.75

At 99% confidence interval estimate of the population mean is,

- E < < + E

155-3.75 < < 155+3.75

151.25< < 158.75

upper confidence interval=158.75

lower confidence interval=151.25

orchestra answered 1 year ago

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