Question

# Question 1 A researcher wishes to find a 90% confidence interval estimate for an unknown population...

Question 1

A researcher wishes to find a 90% confidence interval estimate for an unknown population mean using a sample of size 25. The population standard deviation is 7.2.

The confidence factor z α 2for this est

1.28

1.645

1.96

Question 2

A 90% confidence interval estimate for an unknown population mean μ is (25.81, 29.51).

The length of this CI estimate is

1.96

1.85

3.7

1.645

Question 3

Data below refers to the waiting time in minutes for a sample of 10 customers for an oil change at a certain oil change station. It follows that the time for an oil change for the station follows a normal distribution with population standard deviation σ = 7.5 minutes.

38        42        47        45        35        45        37        34        33        41

Find a 90% CI of unknown mean time μ for an oil change.

[35.8, 43.6]

[23.4, 31.5]

[37.2, 46.1]

Question 4

A 85% confidence interval for the mean time, in minutes, spent by the employees of a company to get to the work is found to be (25 hours, 45 hours). The length for this 85% CI for the mean time is

20 hours

10 hours

70 hours

1)

Given CI level is 90%, hence α = 1 - 0.9 = 0.1
α/2 = 0.1/2 = 0.05, Zc = Z(α/2) = 1.6449

2)

Length = 29.51 - 25.81
= 3.7

3)

sample mean, xbar = 39.7
sample standard deviation, σ = 7.5
sample size, n = 10

Given CI level is 90%, hence α = 1 - 0.9 = 0.1
α/2 = 0.1/2 = 0.05, Zc = Z(α/2) = 1.6449

ME = zc * σ/sqrt(n)
ME = 1.6449 * 7.5/sqrt(10)
ME = 3.9

CI = (xbar - Zc * s/sqrt(n) , xbar + Zc * s/sqrt(n))
CI = (39.7 - 1.6449 * 7.5/sqrt(10) , 39.7 + 1.6449 * 7.5/sqrt(10))
CI = (35.8 , 43.6)

4)

length = 45 - 25 = 20hours

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