Question

1.In a random sample of 23 ​people, the mean commute time to work was 31.1 minutes and the standard deviation was 7.2 minutes. Assume the population is normally distributed and use a​ t-distribution to construct a 80​% confidence interval for the population mean mu. What is the margin of error of mu​? Interpret the results. The confidence interval for the population mean mu is left parenthesis nothing comma nothing right parenthesis . ​(Round to one decimal place as​ needed.)

2.In a random sample of 13 microwave​ ovens, the mean repair cost was ​$80.00 and the standard deviation was ​$15.20. Using the standard normal distribution with the appropriate calculations for a standard deviation that is​ known, assume the population is normally​ distributed, find the margin of error and construct a 95​% confidence interval for the population mean mu. A 95​% confidence interval using the​ t-distribution was left parenthesis 70.8 comma 89.2 right parenthesis. Compare the results.

Answer 1

1). For the given details with population standard deviation unknown we will be using t-distribution for confidence interval calculation as:

μ = M ± t(s_M)

where:

M = sample mean
t = t statistic determined by confidence level and at the degree of freedom=n-1 using T table shown below
s_M = standard error = √(s²/n)

the margin of error is calculated as:

=t(s_M)

=1.32*1.5

=1.984

Interpretation:

The margin of error is the maximum allowable error for the mean commute time to work of all people.

The Confidence interval as:

M = 31.1
t = 1.32 computed using T table at 23-2=22 degree of freedom
s_M = √(7.2²/23) = 1.5

μ = M ± t(s_M)
μ = 31.1 ± 1.32*1.5
μ = 31.1 ± 1.984

Hence 80% CI [29.1, 33.1]

2). Here with thee giveen details and known population standard deviation the confidence interval is caluculated as:

μ = M ± Z(s_M)

where:

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

The Margin of error is calculated as:

Z(s_M)

=1.96*4.22

=8.26

and the confidence interval as:

M = 80
Z = 1.96 9 Calculated using Z table shown below or can be computed using excel )
s_M = √(15.2²/13) = 4.22

μ = M ± Z(s_M)
μ = 80 ± 1.96*4.22
μ = 80 ± 8.26

95% CI [71.7, 88.3].

Since the Confidence interval using t-distribution is (70.8, 89.2) which we can say that it is wider than the confidence interval calculated using Z distribution.

The Z table and t table are: