I need the work shown for solving the problems to compare it with my own work. I

Question

I need the work shown for solving the problems to compare it with my own work. I also use a TI-84 Plus calculator if a formula is needed, Thanks

12) The National Center for Educational Statistics surveyed 4400 college graduates about the lengths of time required to earn their bachelors degrees. The mean is 5.15 years and the standard deviation is 1.68 years. Based on this sample, construct a 90% confidence interval for the mean time required by all college graduates a) 5.08-5.22 5.1-5.19 c) 4.01-6.39 d) 4.37-6.06 13) Which statistic would be more informative about gas mileage for a single person buying a new car? a) point estimate interval estimate c) confidence interval d) sampling distribution 14) What is the degrees of freedom given a sample size of 22? b) 2321 a) 22 d) 20

Explanation / Answer

Solution

Q12

Let X = length of time required to earn bachelor degree. Then, we assume X ~ N(µ, 2), where

µ = population mean and = population standard deviation.

100(1 – ) % confidence interval for µ when 2 is unknown is: {Xbar ± (s/n)(t/2)}, where

Xbar = sample mean,

s = sample standard deviation,

n = sample size and

t/2 = upper (/2) % point of t-Distribution with (n - 1) degrees of freedom..

Given, n = 4400, = 0.1, Xbar = 5.15, s = 1.68, and t/2= t4399, 0.052= 1.645, [using Excel Function],

99% Confidence Interval for µ is: {5.15 ± (1.68/4400)(1.645)} = 5.15 ± 0.041688

Lower Bound = 5.108332, Upper Bound = 5.191668 ANSWER Option (b)

Q13

Answer Option (c)

Confidence Interval is a preferred option to Interval Estimate because the latter is a general term whereas the former also specifies the confidence level

Q14

In general, degrees of freedom = sample size – 1. So, the answer is 22 – 1 = 22. Option (c)

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