In: Statistics and Probability
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.
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 Ha: 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.