# Calculating probability of Type II Error for a one-sample t test [closed]

So I have the following problem:

A transportation company is suspicious of the claim that the average useful life of certain tires is at least 28,000 miles. To verify that, 40 tires are placed in trucks and an average useful life of 27463 is obtained with a standard deviation of 1348 miles.

a) Test this hypothesis with a level of significance of α = 0.01

b) If $\mu_1 = 27230,$ calculate the probability of Type II Error

Any hint on how to do b)?

• Hey I would reply in an answer but I am on mobile... this link should help. ssc.wisc.edu/~gwallace/PA_818/Resources/… Just scroll to the very bottom – Brandon Dec 3 '16 at 2:31
• This is for part b – Brandon Dec 3 '16 at 2:47
• For part a just do a t-test – Brandon Dec 3 '16 at 2:47
• That was definitely helpful, thank you! However, I still don't how the value of σ. I have no idea how to get it – AC Zepp Dec 3 '16 at 3:25
• Use what they gave you. 1348 – Brandon Dec 3 '16 at 3:26

Did you figure out this problem?

Part A) Use a T-test in the graphing calculator

the t-statistic is given by $$t = \frac{\left.\left(\bar{X}-\mu \right.\right)}{\frac{s}{\sqrt{n}}}$$

thus,

$$t = -2.5195$$

For a significance level of $.01$, the t-critical value is about $2.425$

If |t| > t-critical, you can declare statistical significance and reject the null hypothesis

The null hypothesis is $H_0: \mu_ \ge \mu_0,$ and the alternative hypothesis is $H_a: \mu_ < \mu_0,$

Now to find the actual p-value you can use your graphing calculator

Using T-test in graphing calculator: $$P-value = .00798$$ Thus, we have sufficient evidence that the claim "the average useful life of certain tires is at least 28,000 miles" is invalid, it is actually less. $OR$ we have insufficient evidence to conclude that the claim "the average useful life of certain tires is at least 28,000 miles" valid.

Part B Update, This is what I think it is, based on your most recent comment $$2.326*\frac{1348}{\sqrt{40}} = 495.758$$ and $$28000 - 495.758 = 27504.2$$

Now,

$$\frac{27504.2\, -27230}{\frac{1348}{\sqrt{40}}} =1.286$$

Type 2 Error = $$1 -P(z > 1.286)$$

Thus, $$1 - .9007785 = .09922$$

OR $$P(z \leq 1.286) = .09922$$

• Wait I followed the steps given in: ssc.wisc.edu/~gwallace/PA_818/Resources/… And I came up with: Xcritical = (-2.52)(1348) + 28000 = 24603.04 P((z < 27463−28000)/1348) = P(z < -1.95) 0.9941 = 99.41% – AC Zepp Dec 5 '16 at 2:18
• Ahh I see. Yeah that actually looks good. Do you know if that is right? Like can you check in the back of your textbook or no?? I will get rid of my answer for part B. – Brandon Dec 5 '16 at 2:27
• It isn't on any textbook but tomorrow I'll talk to my professor to figure out the answer & I'll make sure to post it here. Thank you so much for your help! – AC Zepp Dec 5 '16 at 2:34
• For one-sided t test in Minitab, I got $T = -2.52$ and $P-value 0.008.$ To get the power of a t test, you need to use the noncentral t distribution. For a power computation you'd need to make a guess at pop SD $\sigma.$ Usual to use sample SD $S$ as an estimate of $\sigma$ for lack of better info. Power is 1 - P(Type II Err). Many software packages have a programmed procedure for finding power. – BruceET Dec 5 '16 at 8:49
• So it might be a little late but I did ended up figuring out this problem, I think. b) Xcritical = ((-2.326)(1348/√40)) + 28000 = 27504.24214, then z > (27504.24214 - 27230) / (1348/√40), equals 1 - (z<1.29), which is 1 - (z<0.9015) = 0.0985, so P(Type II Error) = 9.85%, at least I think that's it – AC Zepp Dec 12 '16 at 1:05

Test of hypothesis: Testing $H_0: \mu = 28000$ vs $H_a: \mu < 28000,$ based on $n = 40$ observations with $\bar X = 27463$ and $S = 1348,$ we Reject $H_0$ at level 1% because the P-value is less than 1%.

Here are results from Minitab statistical software.

One-Sample T

Test of μ = 28000 vs < 28000

N   Mean  StDev  SE Mean  99% Upper Bound      T      P
40  27463   1348      213            27980  -2.52  0.008


Obviously, 27463 < 28000. The question is whether it is enough smaller that we shouldn't ascribe the difference to random variation. The answer (from the one-sided 99& CI) is that any sample mean below 27980 would be significantly smaller.

Power computation for specified alternative. Minitab's 'power and sample size' procedure uses the noncentral t distribution to find power given $n,$ the direction of the test, the difference between null and alternative values, the significance level, and (an estimate $\hat \sigma = 1348$ of) of $\sigma.$ The Type II Error probability is 1 - Power, so the probability of Type II error is about 0.123.

   1-Sample t Test

Testing mean = null (versus < null)
Calculating power for mean = null + difference
α = 0.01  Assumed standard deviation = 1348

Sample
Difference    Size     Power
-770      40  0.876884


Power computations require a population variance, or at least the speculation of one. Usual practice is to use a sample variance as an estimate if one is available. If the true mean endurance of a tire is as low as 27,230, using data from 40 tires gives us a pretty good (approx. 88%) chance of detecting that the claimed 28,000 is an exaggeration. (Notice that the difference between $\mu_0$ and this $\mu_1$ is more than a couple of standard errors.)

Notes: (1) Depending on the level of your class, the intention of the exercise may have been to treat this as a power computation for a z test because $n > 30.$ However, the 'rule' that t and z tests are 'essentially the same' for $n > 30$ is based on the fact that the critical value for a one-sided test at the 5% level $t^*$ is close to $z* = 1.645.$ However, for a test at the 1% level the critical value $t_{.01,39} \approx 2.43$ may not be sufficiently close to the critical value $z_{.01} \approx 2.33.$

(2) In order to do an accurate power computation for a t test, it is necessary to use the noncentral t distribution, as in Minitab. Many mathematical statistics texts have explanations of the noncentral t distribution and its use in power computations. I found the Wikipedia article to be unnecessarily technical, but Section 5 of this paper by Scholz may be helpful.