Question 297954
Part a.)
Start with your significance level.  The standard is usually a 5% significance level.

Null Hypothesis is:  H_ :  u >= 26
Alternate Hypothesis:   H_a : u < 26

Part b.)
I am going to assume this sample set follows a Normal Distribution.  One with a normal density function.


we want to find  P(u >= 26) = 100% - 5%

Standardize this normal distribution so we can look it up in the Z-scores table.
  Z = (x - 26)/s , where Z is a random variable with a standard normal distribution. and s is the standard deviation.  s = 4.83
For a particular case we have Z = (25.02 - 26)/4.83 = 0.202898

Part c)  To test the hypothesis  I found P(Z >= 0.20289) which is our P-value.  We have to compare the P-value to our significance level.  and since the P-value here is greater than 5%, I cannot reject the null hypothesis.

Part d)  The P-value means the probability that the average of the miles per gallon variable is at least 26 mpg.  In this cas the probability is 41.3% 

Part e)
In the Z-scores table this corresponds to P(Z >= 0.20289) = 0.4129
Which is 41.3% > 5%,  Z is not in the rejection region, so therefore I cannot reject the null hypothesis.  I would have to conclude that the average mpg is 26 miles per gallon with a 95% confidence interval.





Sincerely Alan C, 
CSU Math Graduate 07