statistical math question?

Question

The manufacturer of the X-15 steel-belted radial truck tire claims that the mean mileage the tire can be driven before the tread wears out is 60,000 miles. The standard deviation of the mileage is 5,000 miles. The Crosset Truck Company bought 48 tires and found that the mean mileage for their trucks is 59,500 miles. Is Crosset's experience different from that claimed by the manufacturer at the .05 significance level?

Answer 1

first set up the hypothesis

H0: μ = 60000 vs. H1: μ ≠ 60000

Hypothesis Test for mean:

assuming you have a large enough sample such that the central limit theorem holds and the mean is normally distributed then to test the null hypothesis H0: μ = Δ

find the test statistic z = (xBar - Δ) / (sx / sqrt(n))

where xbar is the sample average
sx is the sample standard deviation
n is the sample size

The p-value of the test is the area under the normal curve that is in agreement with the alternate hypothesis.

H1: μ > Δ; p-value is the area to the right of z
H1: μ < Δ; p-value is the area to the left of z
H1: μ ≠ Δ; p-value is the area in the tails greater than |z|


the test statistic is:

z = (59500 - 60000) / (5000 / sqrt(48))
= -0.6928203

the p-value of the test is:

P( |Z| > 0.6928203) = 0.4884223

with such a large p-value we conclude the null hypothesis is plausible. Crosset's experience is not unexpectedly different.


/// ==== notes on picking the hypothesis

Consider the hypothesis as a trial against the null hypothesis. the data is evidence against the mean. you assume the mean is true and try to prove that it is not true.

If the question statement asks you to determine if there is a difference between the statistic and a value, then you have a two tail test, the null hypothesis, for example, would be μ = d vs the alternate hypothesis μ ≠ d

if the question ask to test for an inequality you make sure that your results will be worth while. for example. say you have a steel bar that will be used in a construction project. if the bar can support a load of 100,000 psi then you'll use the bar, if it cannot then you will not use the bar.

if the null was μ ≥ 100,000 vs the alternate μ < 100,000 then will will have a meaningless test. in this case if you reject the null hypothesis you will conclude that the alternate hypothesis is true and the mean load the bar can support is less than 100,000 psi and you will not be able to use the bar. However, if you fail to reject the null then you will conclude it is plausible the mean is greater than or equal to 100,000. You cannot ever conclude that the null is true. as a result you should not use the bar because you do not have proof that the mean strength is high enough.

if the null was μ ≤ 100,000 vs. the alternate μ > 100,000 and you reject the null then you conclude the alternate is true and the bar is strong enough; if you fail to reject it is plausible the bar is not strong enough, so you don't use it. in this case you have a meaningful result.

Any time you are defining the hypothesis test you need to consider whether or not the results will be meaningful.

Answer 2

The answer is yes Crosset's experience is different from that claimed by the manufacturer at the 5% confidence level

Assuming normal distribution then z score (x-u)/sigma for the above is (59500-60000)/5000 = -0.1 and the confidence level for this would be:

normcdf(z,0,1) - normcdf(-z,0,1) = 0.08 = 8%

so 92% of the time the mean mileage calculation would be somewhere between 59500 and 60500 miles, which higher than the 5% level

Answer 3

mean is average and standard deviation is a measure of the spread of its values.