Question

An industrial expert claims that the average useful lifetime of a typical car transimssion which comes with ten years warranty is significantly more than 10 years. In order to test this claim, 9 car transmissions are randomly selected and their useful lifetimes are recorded. The sample mean lifetime is 13.5 years and the sample standard deviation is 3.2 years. Assuming that the useful lifetime of a typical car transmission has a normal distribution, based on these sample result, the correct conclusion at 1% significance level for this testing hypotheses problem is: Group of answer choices

Answer 1

Explanation:

The question is incomplete. The missing information is the group of answer choices. The group of answer choices are

a) none of the above

b) Data provides sufficient evidence, at 1% significance level, to reject the expert's claim. In addition the p-value (or the observed significance level) is equal to P( T < -3.281).

c) Data provides insufficient evidence, at 1% significance level, to support the expert's claim. In addition the p-value (or the observed significance level) is equal to P( Z > 2.896).

d) Data provides sufficient evidence, at 1% significance level, to support the expert's claim. In addition the p-value (or the observed significance level) is equal to P( T >3.355).

e) Data provides insufficient evidence, at 1% significance level, to support the researcher's claim. In addition the p-value (or the observed significance level) is equal to P(Z > 2.896).

Solution:

We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean

For the null hypothesis,

H0: µ ≥ 10

For the alternative hypothesis,

µ < 10

This is a left tailed test.

Since the number of samples is small and the population standard deviation is not given, the distribution is a student's t.

Since n = 9,

Degrees of freedom, df = n - 1 = 9 - 1 = 8

t = (x - µ)/(s/√n)

Where

x = sample mean = 13.5

µ = population mean = 10

s = samples standard deviation = 3.2

t = (13.5 - 10)/(3.2/√9) = 3.28

Since α = 0.01, the critical value is determined from the t distribution table. Recall that this is a left tailed test. Therefore, we would find the critical value corresponding to 1 - α and reject the null hypothesis if the test statistic is less than the negative of the table value.

1 - α = 1 - 0.01 = 0.99

The negative critical value is - 2.896

Since - 3.28 is lesser than - 2.896, then we would reject the null hypothesis.

By using probability value,

We would determine the p value using the t test calculator. It becomes

p = 0.0056

Level of significance = 1%

Since alpha, 0.01 > than the p value, 0.0056, then we would reject the null hypothesis. Therefore, At a 1% level of significance, the sample data showed significant evidence that the average useful lifetime of a typical car transimssion which comes with ten years warranty is significantly less than 10 years

The correct option is

a) none of the above