~Class Meeting #22---Testing a claim about a mean
(Population standard deviation is known)

~Note:  Again, not a very realistic situation since knowing the population standard deviation without knowing the population mean usually does not occur.  

~Assumptions:  
1) Population standard deviation is known  
 
2) Either n ≥30 or the population is normally distributed
   (either one or both)

~Test statistic:   z = ( x bar - m_o) / [s/√n],  x bar is the sample mean
and m_o comes from the null hypothesis

~p-values:  use the TI-83 or table A-2 (standard normal curve).

~Critical values:  use the TI-83 or table A-2 (standard normal curve).

~Note:  There are two methods you can use.  The standard method (using  critical regions) or the p-value method (TI-83).

~The p-value method is recommended since the TI-83 does all the work for you.  However, you must understand  both.

~Example:  In 1999, the average retail price of a certain line of books was $24.96.  A sample of 40 such books gave an average retail price of $26.10 in today’s market. Test the claim that the average retail price has not increased from the 1999 average at the .10 level of significance. Assume the standard deviation of the price of this line of books is estimated to be $8.33. Use both methods. State the conclusion properly.

~Method 1: (standard method)

  H_o: m = 24.96           a = .10
  H₁: m > 24.96           To get the critical value, Z, use 2nd Vars
                                       menu 3, Invnorm(.90) = 1.28
                                       s = 8.33

  Test statistic = z = (26.10-24.96)/[8.33/√40] = .866

   Since the test statistic is not in the rejection region, we do not reject H_o.

     Conclusion: Based on our sample of 40 books, we have insufficient evidence to conclude that the average retail price of these books has increased. ( at the .10 level of significance)


~Method 2: (p-value method using the TI-83)

  Use STAT, TESTS, Z-Test (menu 1), stats, m_o: 24.96, s: 8.33, sample mean: 26.10, n: 40,  > m_o, calculate, enter.

  Look for p = .19336972 which is not less than .10, so we do not reject the null hypothesis.  Conclusion is stated the same as above.

~Example:  A company that produces snack foods uses a packaging machine to package 454-gram bags of  chips. The quality control department sampled 50 bags and found the average to be 451.21 grams with a population standard deviation of 7.9 grams.

Test, at the 5% level, whether the sample data provides sufficient evidence that the packaging machine is not working properly.

Ans: (use the TI-83),  H_o: m = 454,  H₁: m ≠ 454
        (this is a two-tailed test),  a = .05,  s = 7.9,  n = 50,
        p-value is .0125 < .05, so, reject Ho.

Conclusion: There is sufficient evidence to conclude that the packaging machine is not working properly.