Question

1. A simple random sample of size n equals 350 individuals who are currently employed is asked if they work at home at least once per week. Of the 350 employed individuals​ surveyed, 32 responded that they did work at home at least once per week. Construct a​ 99% confidence interval for the population proportion of employed individuals who work at home at least once per week.

QUESTION- what is the lower and upper bound

2. A simple random sample of size n=21 is drawn from a population that is normally distributed. The sample mean is found to be x "overbar" = 51 and the sample standard deviation is found to be equals=10. Construct a 95​% confidence interval about the population mean.

QUESTION- what is the lower and upper bound

3. A simple random sample of size n= 40 is drawn from a population. The sample mean is found to be x "overbar" equals 120.4 and the sample standard deviation is found to be s= 12.8. Construct a​ 99% confidence interval for the population mean.

QUESTION- what is the lower and upper bound

Answer 1

Solution :

1 ) Given that

n = 350

x =32

= x / n = 32 / 350 = 0.091

1 - = 1 - 0.091 = 0.909

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z/2 = Z0.005 = 2.580

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

= 2.580 * (((0.091 * 0.909) / 350 )

= 0.040

A 90 % confidence interval for population proportion p is ,

- E < P < + E

0.091 - 0.040 < p < 0.091 + 0.040

0.051 < p < 0.131

2 )Given that,

= 51

s = 10

n = 21

Degrees of freedom = df = n - 1 = 21 - 1 = 20

At 95% confidence level the t is ,

  = 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t /2,df = t0.025,20 =2.086

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

= 2.086 * (10 / 21)

= 4.55

Margin of error = 4.55

The 95% confidence interval estimate of the population mean is,

- E < < + E

51 - 4.55 < < 51 + 4.55

46.45 < < 55.55

The lower bound = 46.45

The upper bound = 5.55

3 ) Given that,

= 120.4

s = 12.8

n = 40

Degrees of freedom = df = n - 1 = 40 - 1 = 39

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

t /2,df = t0.005,39 =2.708

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

= 2.708 * (12.8 / 40)

= 5.48

Margin of error =5.48

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

- E < < + E

120.4 - 5.48 < < 120.4 + 5.48

114.92 < < 125.88

The lower bound = 114.92

The upper bound = 125.88