Can you show me how to solve this? Let the following sample of 8 observations be

Question

Can you show me how to solve this?

Let the following sample of 8 observations be drawn from a normal population with unknown mean and standard deviation: 15, 25, 12, 18, 28, 17, 14, 24. Use Table 2.

A) Sample Mean= 19.125 Sample Standard Deviation= 5.82

D) As the confidence level increases, the margin of error becomes larger.

Need help with B and C please

a. Calculate the sample mean and the sample standard deviation. (Round intermediate calculations to 4 decimal places. Round "Sample mean" to 3 decimal places and "Sample standard deviation" to 2 decimal places.)

b. Construct the 95% confidence interval for the population mean. (Round "t" value to 3 decimal places and final answers to 2 decimal places.)

Confidence interval             to

c. Construct the 99% confidence interval for the population mean. (Round "t" value to 3 decimal places and final answers to 2 decimal places.)

Confidence interval             to

d. What happens to the margin of error as the confidence level increases from 95% to 99%?

Explanation / Answer

The given dataset is of 8 observations.

15, 25, 12, 18, 28, 17, 14, 24.

Here sample mean (Xbar) = 19.125

sample standard deviation (s) = 5.82

Here we use t-interval because population standard deviation is unknown and sample size is small.

The confidence interval for population mean (mu) is,

Xbar - E < mu < Xbar + E

where Xbar is sample mean.

E is the margin of error.

E = (tc*s) / sqrt(n)

tc is the critical value for t-distribution and

s is sample standard deviation.

tc we can find by using EXCEL.

syntax :

=TINV(probability, deg_freedom)

where probability = 1 - C

deg_freedm = n-1 = 8-1 = 7

b. Construct the 95% confidence interval for the population mean.

Here C = confidence level = 95% = 0.95

tc = 2.365

E = (2.365*5.82) / sqrt(8) = 4.87

Lower limit = Xbar - E = 19.125 - 4.87 = 14.26

Upper limit = Xbar + E = 19.125 + 4.87 = 23.99

The 95% confidence interval for the population mean is (14.26, 23.99)

c. Construct the 99% confidence interval for the population mean.

Here C = confidence level = 99% = 0.99

tc = 3.499

E = (2.365*5.82) / sqrt(8) = 7.20

Lower limit = Xbar - E = 19.125 - 7.20 = 11.92

Upper limit = Xbar + E = 19.125 + 7.20 = 26.33

The 95% confidence interval for the population mean is (11.92, 26.33)

d. What happens to the margin of error as the confidence level increases from 95% to 99%?

As the confidence level increases, the margin of error becomes larger.

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