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. viking truck company bought 48 tires and found that the mean
mileage for their trucks is 59,500 miles. is viking's experience different from that claimed by
the manufacturer at the 0.05 significance level?
(a) state the null hypothesis and the alternative hypothesis. (b) state the decision rule.
(b) compute the value of the test statistic. (d) what is your decision regarding h0 ?
(e) what is the p-value? interpret it.

A)  H0:  µ = 60,000, Ha:  µ ≠ 60,000

B) Read below

C) -0.01

D)  Fail to reject the null hypothesis

e) 0.4960

Step-by-step explanation:

a:  The manufacturer claims that the mean mileage IS 60,000.  Is means equals in math.  That's the claim, so a not equals to sign goes in the alternate hypothesis because if the amount is significantly more or significantly less, then the claim is not correct.  

b: This is a 2 tailed test.  At a 0.05 significance level with a sample size of 48, our critical values are z < -1.96 and z > 1.96.  If our test statistic falls in either of these ranges, then we will reject the null hypothesis

c:  z = (59,500 - 60,000)/(5,000/√48) = -0.01

d:  -1.96 < -0.01 < 1.96, we fail to reject the null hypothesis

e:  P(z < -0.01) = 0.4960.  This means that a sample with these results can be expected to happen about 49.6% of the time

its kinda hard to answer this when we dont have all the information that we need.