Question

1) You measure 42 textbooks' weights, and find they have a mean weight of 64 ounces. Assume the population standard deviation is 9.6 ounces. Based on this, construct a 99% confidence interval for the true population mean textbook weight. Give your answers as decimals, to two places.

__< μ < __

2) Karen wants to advertise how many chocolate chips are in each Big Chip cookie at her bakery. She randomly selects a sample of 70 cookies and finds that the number of chocolate chips per cookie in the sample has a mean of 17.2 and a standard deviation of 3.9. What is the 80% confidence interval for the number of chocolate chips per cookie for Big Chip cookies? Enter your answers accurate to one decimal place (because the sample statistics are reported accurate to one decimal place).
__ < μ < __

3) 44% of 84 delegates in a policitcal convention favored changing the rules to restrict the number of potential candidates. What is the 80% confidence interval for the population proportion.
Give your answers as decimals, to two places.

__ < p < __

Answer 1

Solution

1 ) Given that,

= 64

= 9.6

n = 42

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* (/n)

= 2.576 * (9.6 / 42 )

= 3.81

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

- E < < + E

64 - 3.81 < < 64 + 3.81

60.19 < < 67.81

(60.19 , 67.81 )

2 ) Given that,

= 17.2

s = 3.9

n = 70

Degrees of freedom = df = n - 1 = 70 - 1 =69

At 80% confidence level the t is ,

= 1 - 80% = 1 - 0.80 = 0.20

/ 2 = 0.20 / 2 = 0.10

t _/2,df = t_0.10,69 =1.294

Margin of error = E = t_/2,df * (s /n)

= 1.294 * (3.9 / 70)

= 0.5

Margin of error = 0.5

The 80% confidence interval estimate of the population mean is,

- E < < + E

17.2 - 0.5 < < 17.2 + 0.5

16.7 < < 17.7

(16.7, 17.7)

3 ) Given that,

n = 84

= 0.44

1 - = 1 - 0.44 = 0.56

At 80% confidence level the z is ,

= 1 - 80% = 1 - 0.80 = 0.20

/ 2 = 0.20 / 2 = 0.10

Z_/2 = Z _0.10 = 1.280

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 1.280 * (((0.44 * 0.56) / 84) .

= 0.07

A 80% confidence interval for population proportion p is ,

- E < P < + E

0.44 - 0.07 < p < 0.44 + 0.07

0.36 < p < 0.51