Question

A battery pack used in a medical device needs to be recharged about every 5 hours. A random sample of 20 battery packs is selected and subjected to a life test. The average life of these batteries is 5.05 hours with standard deviation ? = 0.3 hours. Assume that battery life is normally distributed. Is there evidence to support the claim that mean battery life is more than 5 hours? Use ? = 0.01.

a. Write the appropriate hypothesis.

b. Use P-value approach to test the hypothesis.

c. Use t-test to test the hypothesis.

d. Use confidence interval to test the hypothesis.

e. What is your conclusion?

f. If the true mean life is 5.1 hours, what is the type II error?

Answer 1

We want to test for the population mean. But we have not been given the population SD and the data follows normal distribution therefore we can use the t-test.

The claim to test is if the mean life is more than 5 hours. This right one tailed test.

= 5.05 Sx = 0.3 n = 20  ? = 0.01

a. Write the appropriate hypothesis.

(the avg. life is not more than 5 hours)

(the avg. life is more than 5 hours)

b. Use P-value approach to test the hypothesis.

df = n -1 = 19

p-value = P( > |Test Stat|)

c. Use t-test to test the hypothesis.

Test Stat =

Where the null mean = 5

Test Stat = 0.7454

p-value = P( > |Test Stat|)

= P( > 0.75)

p-value = 0.2326

Since p-value > 0.01

We do not reject the null hypothesis at 1%. There is insufficient evidence to support that the avg. life is more than 5 hours.

d. Use confidence interval to test the hypothesis.

(1- )% is the confidence interval for population mean

Here it is right tailed so we only find the lower limit for the test at  ? = 0.01

Interval :

C.V. at  ? = 0.01 =

=

= 2.5395 .........................found using t-dist tables

Interval (4.8796, )

Since this interval contains the null value 5 we do not reject the null hypothesis.

we can see if we conduct the test using same level and same statistics we reach at same result.

e. What is your conclusion?

We do not reject the null hypothesis at 1%. There is insufficient evidence to support that the avg. life is more than 5 hours.

f. If the true mean life is 5.1 hours, what is the type II error?

Type 2 error : Accepted false null hypothesis.

We accept the null if

|test Stat | < C.V.

C.V. at 0.01 for one tail = 2.5395

< 2.5395

Substituting all the values except sample mean

< 5.1704

Type 2 error probability = P( < 5.17) given = 5.1)

= P( < )

= 1- P( > 1.05)

= 1 - 0.15372

Ans: 0.8463