Answer:
Step-by-step 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
