Question

A) Find the z critical value for a 90% confidence level:

Zc=

B) Find the z critical value for an 84% confidence level:

Zc=

C) The higher the confidence, the narrower the confidence interval

True or False

D) The greater the sample size, the mnarrower the confidence interval

True or False

E) In a recent study of 84 eighth graders, the mean number of hours per week that they watched TV was 22.3. Assume population standard deviation is 5.8 hours.

*State the critical value for a 90% confidence level:_________________

*Zc=

*Construct a 90% confidence interal for the mean number of hours of televion watched per week by eighth graders. State the equation of the confidence intercal, then show all values substituted into the equation, find the confidence interval.

*Write a sentence interpreting your results in the context of the problem.

*Construct a 99% confidence interval for the same problem

*Which o the confidence intervals do you think is better, and why?

Answer 1

A) The Z critical values for 90% confidence interval is 1.645

B) The Z critical values at 84% confidence level is 1.401

C) False, because Z value increases as an increase in the confidence interval

D) true, because the Sample size is inversely proportional to the Confidence interval

E)

The formula for estimation is:

μ = M ± Z(s_M)

where:

M = sample mean
Z = Z statistic determined by confidence level
s_M = standard error = √(s²/n)

Calculation

M = 22.3
Z_c= 1.645
s_M = √(5.8²/84) = 0.63

μ = M ± Z(s_M)
μ = 22.3 ± 1.64*0.63
μ = 22.3 ± 1.036

CI= 21.264 to 23.336

Interpretation:

You can be 90% confident that the population mean (μ) falls between 21.264 and 23.336.

Again Similarly

Calculation

M = 22.3
t = 2.58
s_M = √(5.8²/84) = 0.63

μ = M ± Z(s_M)
μ = 22.3 ± 2.58*0.63
μ = 22.3 ± 1.63

You can be 99% confident that the population mean (μ) falls between 20.67 and 23.93.

The 99% Confidence interval is better because it has wider interval range.