Given a sample size of n = 256. Let the variance of the population be 2 = 8.41.

Question

Given a sample size of n = 256. Let the variance of the population be 2 = 8.41. Let the mean of the sample be xbar = 4. Construct a 95% confidence interval for µ, the mean of the population, using this data and the central limit theorem.

a. What is the standard deviation of the population?

b. What is the standard deviation of means of sample size n, xbar, i.e. xbar , in terms of and n?

c. Is this a one-sided or two-sided problem?

d. What value of z should be used in computing k, the margin of error, where z = k/xbar ? e. What is k?

f. Write the 95% confidence interval for µ based on xbar and k, (xbar – k) < µ < (xbar + k) g. Using the Z-score applet “Area from a value”. Let the Mean = 6, and SD = xbar. Choose “Between (xbar -k) and (xbar + k)”. Hit “Recalculate”. Does the probability approximately equal 0.95? (yes or no). Include a screen shot of your answer.

Explanation / Answer

sol:

n=256

n>30

2 = 8.41.

standard deviation of the population

standard deviation =sqrt(8.41)

=2.9

b. What is the standard deviation of means of sample

=standard deviation/sqrt(sample sizE)

=2.9/sqrt(4)

=2.9/2=1.45

Solutionc:

two-sided problem

Solutiond:

z alpha/2 f for 95%=1.96

z=1.96/1.45

z=1.352

k=1.96

95% confidence interval for µ based on xbar and k

=4-1.96(1.45) and 4+1.96(1.45)

=1.158 and 6.842

we are 95% confident that the true population mean lies in between

1.158 and 6.842

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