Question

(PLEASE, SINCE THE VERY BEGINNING, ALL THE ONE BY ONE STEPS NEED TO BE MENTIONED IN YOUR CALCULATION) A manufacturer of cell phones guarantees that his cell phones will last, on average, 3 years with a standard deviation of 1 year. If 5 of those cell phones are found to have lifetimes of 1.9, 2.4, 3.0, 3.5 and 4.2 years, can the manufacturer still be convinced that his cell phones have a standard deviation of 1 year? Test at a 0.05 level of confidence. Thank you in advance for your help!

Answer 1

First we calculate the standard deviation for the sample.

Standard deviation = Sqrt(Variance), where

Variance = Sum of Squares/n -1 = Sum( X - Mean)²/n-1

#	X	Mean	(x - mean)2
1	1.9	3	1.21
2	2.4	3	0.36
3	3	3	0
4	3.5	3	0.25
5	4.2	3	1.44
Total	15		3.26

n	5
Sum	15
Average	3.000
SS(Sum of squares)	3.26
Variance = SS/n-1	0.815
Std Dev=Sqrt(Variance)	0.9028

Given : = 1

The Hypothesis:

H0: = 1   (The Claim)

Ha: 1

The Test Statistic:

The p value: The p value (2 tailed) for = 3.26, df = n -1 = 4 is; p value = 0.9694

(To find the 2 tailed probability, I have used the excel formula CHISQ.DIST(3.26,4,TRUE), which returns the left tailed probability of 0.4847. The left tailed probability * 2, gives us the 2 tailed probability).

The Decision Rule: If p value is < , Then Reject H0.

The Decision: Since p value (0.9694) is > , we fail to reject H0.

The Conclusion: There is insufficient evidence at the 95% level of significance to conclude that the standard deviation of the life of the cell phones is not 1 year.