Question

3). You are given the claim that the mean of a population is not equal to 24 cm. You
don’t believe in this claim and so you want to test it. Suppose that you know the
population standard deviation is 4 cm, and the population distribution is approximately
normal. To test this claim, you take a random sample as follows
X = (20, 23, 22, 24, 24, 24, 25, 26, 24, 23, 27, 24, 29, 20, 25, 26, 28).
Is there enough evidence to support the claim? Justify your answer completely.

4) Based on problem 3 above, compute the 95 % confidence interval of the true mean.
Interpret this confidence interval and also explain what is the meaning of the 95%
confidence level.

Answer 1

Question 3

Here, we have to use one sample z test for the population mean.

The null and alternative hypotheses are given as below:

H₀: µ = 24 versus H_a: µ ≠ 24

This is a two tailed test.

The test statistic formula is given as below:

Z = (Xbar - µ)/[σ/sqrt(n)]

From given data, we have

µ = 24

Xbar = 24.35294118

σ = 4

n = 17

α = 0.05

Critical value = -1.96 and 1.96

(by using z-table or excel)

Z = (24.35294118 - 24)/[4/sqrt(17)]

Z = 0.3638

P-value = 0.7160

(by using Z-table)

P-value > α = 0.05

So, we do not reject the null hypothesis

There is sufficient evidence to conclude that the mean of a population is not equal to 24 cm.

Question 4

Confidence interval for Population mean is given as below:

Confidence interval = Xbar ± Z*σ/sqrt(n)

From given data, we have

Confidence level = 95%

Critical Z value = 1.96

(by using z-table)

Confidence interval = Xbar ± Z*σ/sqrt(n)

Confidence interval = 24.35 ± 1.96*4/sqrt(17)

Confidence interval = 24.35 ± 1.9014

Lower limit = 24.35 - 1.9014 = 22.45

Upper limit = 24.35 + 1.9014 = 26.25

Confidence interval = (22.45, 26.25)

WE are 95% confident that the population mean will lies between above interval.