John claims that the typical lifetime of a car in his shop with 10 years warranty, is significantly more than 10 years. To test the claim, 9 cars are randomly selected, with lifetimes recorded. Sample mean lifetime is 13.5 years, and sample standard deviation is 3.2 years. Assuming the lifetime has a normal distribution, what conclusion can be drawn at 1% significant level?

So I have several tables and formulas here, first I am looking to see if it's a one sample or two sample problem, it's one sample. Then I am looking to see of the sample is large or small, it's small, only 9. Then I am looking to see if the population variance or standard deviation is known, or unknown, and it's unknown. And lastly I am looking for normality, and it's normally distributed. So with this information, I'd assume we're using the T test looking for the mue (population mean) parameter.

Which then follows:

$H_0: \mu = 10$

$H_a: \mu > 10$

$T = (\bar X - \mu)/(S/\sqrt{n})$

Plugging our values in, I get 3.281 as a T-score.

The rejection region is then defined by $T > T_\alpha(\nu)$ where $\nu = n - 1.$

So on the T-table, with degrees 8, and an alpha of .01, this would be the same as the $T > -T_1-\alpha$ so alpha would be now .99, and we can find the T-score of .99 from the table ensuring to make it negative as -2.896.

Our P-value would also be $P(T > 3.281).$

Now since our test statistic which is 3.281, is greater than -2.896, it's in the rejection region and we can reject the null hypothesis. Which in turn then, we can also support John's claim. So if I did all this correct, great, but the options I am given for this problem are the following:

a. Sufficient evidence at 1% to support the claim, and the p-value is equal to P(T > 3.355).

b. Sufficient evidence at 1% to reject the claim, and the p-value is equal to P(T < -3.281).

c. Insufficient evidence at 1% to support the claim, and the p-value is equal to P(Z > 2.896).

d. Insufficient evidence at 1% to support the claim, and the p-value is equal to P(Z > 2.896).

e. None of these answers

Is the answer here really e? Did I do everything correct? Already seems kind of a funky question as one answer is repeated and we're not even using Z. And the answer could be a. but that t-score is if the alpha level was 0.005, and we have an alpha of 0.01, and it's not a two tail test, it's a one tail test. And then b. is wrong because it's not negative nor <.

  • $\begingroup$ Answer A would only make sense to me if it was a two-tail test, as we'd then be dividing the alpha by 2... $\endgroup$
    – Zion Todd
    Commented Apr 24, 2017 at 17:59

1 Answer 1


This is a one-sample test because you are comparing results of your sample of size $n = 9$ against a hypothetical value (the 10 year warranty length).

It is a t test because the population is assumed to be normal and the population SD $\sigma$ is unknown (estimated by $S$). [Correctly understood, such a test is a t test (not a z test) whenever $\sigma$ is unknown, regardless of the sample size.]

Thus, as you say, you are testing the null hypothesis $H_0: \mu = 10$ against the one-sided alternative $H_a: \mu > 10.$ The test statistic is $$T = \frac{\bar X - \mu_0}{S/\sqrt{n}} = \frac{13.5 - 10}{3.2/\sqrt{9}} = 3.281.$$

Under $H_0$ (jargon for 'assuming the null hypothesis to be true), $T \sim \mathsf{T}(8),$ Student's t distribution with degrees of freedom $n - 1 = 9 - 1 = 8.$

Working at the fixed significance level $\alpha = .01 = 1\%,$ we reject $H_0$ when $T \ge t^* = 2.896,$ where $t^*$ cuts probability 0.01 from the upper tail of $T \sim \mathsf{T}(8).$ You can find $t^* = 2.896$ on the row of a printed t table marked DF = 8 or by using software. (The computation from R statistical software is shown below.) Because $T = 3.281 > t^* = 2.896$ you reject $H_0$ at the 1% level.

qt(.99, 8)
## 2.896459

Generally speaking, you can't find the exact P-value of a t test from printed tables because a table gives you only a few probabilities. Using software, you can find the P-value in this problem as $P(T > 3.281) = 0.0056.$ So you could reject at the 5% level, the 1% level, or even the 0.6% level, but not at the 0.5% level.

1 - pt(3.281, 8)
## 0.005585558

Thus P-values have become popular only in the computer age. Their advantage is that each person can test at their own significance level $\alpha$ one the P-value is known. In the t table in my book the t-values that bracket 3.281 are 2.896 at tail probability 0.01 and 3.355 at tail probability 0.005,, so all I can say about the P-value from looking at this table is that the P-value is between 0.01 and 0.005.

The figure below shows the density function ot $\mathsf{T}(8).$ The vertical blue line is the observed value T = 3.281, and the area under the curve to the right of it is 0.00556. The dotted vertical red line at 2.896 cuts 1% of the probability from the upper tail of this distribution.

enter image description here

Note: As far as your multiple choice questions is concerned, it may be a trick question. Note that $P(T > 3.281) = P(T < - 3.281)$ if you're thinking of the symmetrical random variable $T \sim \mathsf{T}(8)$. But it is strange to write $P(T < - 3.281),$ if you're thinking of $T$ as the observed value of the t statistic. This is one reason I'm not a fan of multiple-choice tests. (Contrary to student lore, there are times when 'none of the above' is the best answer. Not sure if this is one of them.)

  • $\begingroup$ Thank you for yet another superb answer, really helped me understand a lot more than I had asked. I guess the t-value does not turn negative then as I had originally understood, I guess it just means that they are equal T with alpha .99 and T with alpha .01. Which would make sense since it is symmetrical. $\endgroup$
    – Zion Todd
    Commented Apr 24, 2017 at 19:52
  • $\begingroup$ After you Accepted, I extended my note at the end about the multiple choice question. $\endgroup$
    – BruceET
    Commented Apr 24, 2017 at 20:04

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