Question

From your sample of 80 5-year-old girls, you notice that 9 of them have red-hair.

Find and interpert a 99% confidence interval for the proportion of red-haired 5-year-olds.

1. State the confidence interval (round to 4 decimal places):

2. Interpert the confidence interval in the SHOW YOUR WORK area.

3. You were told that 2% of the population is red-haired. Based on your interval, you can conclude that:

A.

the proprotion of red-hair 5-year-olds is lower than 2% since 2% is higher than the interval.

B.

the proportion you were told is correct since 2% is within the interval.

C.

the proportion of red-hair 5-year-olds is higher than 2% since 2% is lower than the interval.

D.

the proportion of red-hair 5-year-olds is higher than 2% since 2% is higher than the sample proportion.

Answer #1

Answer)

Given n (sample size) = 80

And point estimate (p) = 9/80

First we need to check the conditions of normality that is, if n*p and n*(1-p) both are greater than 5 or not

N*p = (80*9/80) = 9

N*(1-p) = 71

As both the conditions are met, we can use standard normal.z table to estimate the interval

From z table, critical value z for 99% confidence level is 2.58

Margin of error (MOE) = Z*√{P*(1-P)/N}

MOE = 0.0911454838088

Confidence interval is given by

P-MOE < P+MOE

0.0213545161911 < P < 0.2036454838088

0.0214 < P < 0.2036

2)

We are 99% confident that true population proportion lies in this interval

3)

As 2% is in the interval

So, option B is correct

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