Question

1. Determine the z critical value for a 92% confidence level. Show work or explain how you found your answer

2. Determine the t critical value for a 95% confidence level when n =15. Show work or explain how you found your answer.

3. Ten bagels were selected at Brueggers. If the sample mean weight was 1.4 oz. with a population standard deviation of 0.4 oz., construct a 90% confidence interval for the population mean. Interpret your answer.

4.To estimate the average GPA of CSCC students, Sara wants to know how large a sample she needs to survey. She believes the standard deviation is 0.52 and wants to construct a 99% confidence interval, with a margin of error no more than 0.10. How many students does she need to survey?

Answer 1

Solution :-

At 92% confidence level

= 1 - 92%  

= 1 - 0.92 = 0.08

/2 = 0.04

Z/2 = Z_0.04 = 1.751

The critical value z = 1.751

2.

sample size = n = 15

Degrees of freedom = df = n - 1 = 15 - 1 = 14

At 95% confidence level

= 1 - 95%

=1 - 0.95 = 0.05

/2 = 0.025

t/2,df = t_0.025,14 = 2.145

The critical value t = 2.145

3.

Given that,

Point estimate = sample mean = = 1.4

Population standard deviation =    = 0.4

Sample size = n = 10

At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z_{0.05 = 1.645}

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

= 1.645 * ( 0.4 /  10 )

= 0.21

At 90% confidence interval estimate of the population mean is,

- E < < + E

1.4 - 0.21 <   < 1.4 + 0.21

1.19 <   < 1.61

( 1.19 , 1.61 )

The 90% confidence interval estimate of the population mean is : ( 1.19 , 1.61 )

4.

Population standard deviation = = 0.52

Margin of error = E = 0.10

At 99% confidence level the z is,

= 1 - 99%

= 1 - 0.99 = 0.01

/2 = 0.005

Z_/2 = Z_0.005 = 2.576

sample size = n = [Z_/2* / E] ²

n = [ 2.576 * 0.52 / 0.10 ]²

n = 179.43

Sample size = n = 180