Question

A manufacturer claims that the mean lifetime of its lithium batteries is 1500 hours. A home owner selects 30 of these batteries and finds the mean lifetime to be 1470 hours with a standard deviation of 80 hours. Test the manufacturer's claim. Use a=0.05 . Round the test statistic to the nearest thousandth.

a) Hypothesis :

b) Critical value (tcritical) :

c) Test statistic (tstat) and the decision about the test statistic: (reject or fail to reject Ho) :

d) Conclusion that results from the decision about Ho, how would you politely address the manufacturer’s claim?

Answer 1

Solution:

One sample t-test

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

The null and alternative hypotheses are given as below:

Null hypothesis: H0: the mean lifetime of its lithium batteries is 1500 hours.

Alternative hypothesis: Ha: the mean lifetime of its lithium batteries is different than1500 hours.

H0: µ = 1500 versus Ha: µ ≠ 1500

This is a two tailed test.

The test statistic formula is given as below:

t = (Xbar - µ)/[S/sqrt(n)]

From given data, we have

µ = 1500

Xbar = 1470

S = 80

n = 30

df = n – 1 = 29

α = 0.05

Critical value = - 2.0449 and 2.0449

(by using t-table or excel)

t = (1470 – 1500)/[80/sqrt(30)]

t = -2.053

Test statistic = -2.053

P-value = 0.0491

(by using t-table)

P-value < α = 0.05

So, we reject the null hypothesis

There is insufficient evidence to conclude that the mean lifetime of its lithium batteries is 1500 hours.

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