QA 233 (Basic Business Statistics)

Solutions to Practice Problem Set IX

1. Finyl Vinyl, the last of the great LP-record producers, claims that the perceived quality of vinyl records and compact disks are equal. They first play a randomly selected song from a vinyl copy of Paula Cole’s ‘This Fire’ for a random sample of forty-two consumers. They then play the same song from a compact disk for the same forty-two consumers. The consumers were then asked to indicate whether they thought the vinyl or compact disk had a better sound quality. Their responses are given below.

   Consumer	Preference	Consumer	Preference	Consumer	Preference
   01	VINYL	15	COMPACT DISC	29	VINYL
   02	COMPACT DISC	16	VINYL	30	COMPACT DISC
   03	COMPACT DISC	17	COMPACT DISC	31	COMPACT DISC
   04	VINYL	18	COMPACT DISC	32	VINYL
   05	COMPACT DISC	19	VINYL	33	VINYL
   06	VINYL	20	COMPACT DISC	34	COMPACT DISC
   07	VINYL	21	COMPACT DISC	35	VINYL
   08	VINYL	22	COMPACT DISC	36	COMPACT DISC
   09	COMPACT DISC	23	COMPACT DISC	37	COMPACT DISC
   10	COMPACT DISC	24	VINYL	38	VINYL
   11	VINYL	25	VINYL	39	COMPACT DISC
   12	COMPACT DISC	26	COMPACT DISC	40	VINYL
   13	COMPACT DISC	27	VINYL	41	COMPACT DISC
   14	COMPACT DISC	28	VINYL	42	VINYL

   Use the data that have been collected to perform the appropriate hypothesis test at the a=.05 level of significance.

   This problem involves testing a hypothesis where our random variable is whether a respondent prefers the sound quality of vinyl recordings over compact discs. Thus we are dealing with a nominal variable and so will be testing a hypothesis about a single proportion. The steps in this hypothesis test are:

   1.  H₀:p = 0.50 where p is the proportion of respondents who prefer the sound quality of vinyl recordings over compact discs.

         H_a:p ¹ 0.50

   2.  Since we are using a large sample (np₀ = 42(0.50) = 21 ³ 30 & n(1-p₀) = 42(1-0.50) = 21 ³ 30) to test a hypothesis about a single proportion, use 

   

   3.    We have a two-tailed test at the a=.05 level of significance, so z_a/2 = ±1.96. Graphically we have:

   

                    The resulting decision rule is:

   If -1.96 £ z £ 1.96 then do not reject H₀.

   Otherwise reject H₀.

   4.  Using our sample data (and noting that 19 out of 42 sample respondents preferred vinyl), we calculate the test statistic:

   

   5.   Implementation of our decision rule

   -1.96 £ -0.617 £ 1.96 so do not reject H₀

   leads us to conclude that the sample results do not support the conclusion that the proportion of consumers who prefer vinyl differs from the proportion of consumers who prefer compact discs.

2. The Quality Control Manager of a potato chip manufacturer believes that at least one-fourth of all 16 oz. bags of his brand are overfilled. He randomly selects thirty bags from recent production runs and records their weight. The results are

   Bag 1	16.0 oz.	Bag 11	16.6 oz.	Bag 21	15.6 oz.
   Bag 2	15.2 oz.	Bag 12	15.9 oz.	Bag 22	15.3 oz.
   Bag 3	15.9 oz.	Bag 13	15.2 oz.	Bag 23	15.7 oz.
   Bag 4	16.7 oz.	Bag 14	15.3 oz.	Bag 24	15.2 oz.
   Bag 5	15.0 oz.	Bag 15	15.7 oz.	Bag 25	15.3 oz.
   Bag 6	15.4 oz.	Bag 16	16.3 oz.	Bag 26	16.9 oz.
   Bag 7	15.3 oz.	Bag 17	15.5 oz.	Bag 27	15.4 oz.
   Bag 8	15.6 oz.	Bag 18	15.7 oz.	Bag 28	15.9 oz.
   Bag 9	15.7 oz.	Bag 19	15.1 oz.	Bag 29	15.6 oz.
   Bag 10	16.4 oz.	Bag 20	15.0 oz.	Bag 30	15.9 oz.

   Use these sample results to test the appropriate hypothesis at an a=.01 level of significance.

   This problem involves testing a hypothesis where our random variable is whether a bag of chips is overfilled. Thus we are dealing with a nominal variable and so will be testing a hypothesis about a single proportion. The steps in this hypothesis test are:

   1.      H₀:p ³ 0.25 where p is the proportion of bags of chips that are overfilled.

            H_a:p < 0.25

   2.     Since we are using a large sample (np₀ = 30(0.25) = 7.5 ³ 30 & n(1-p₀) = 30(1-0.25) = 22.5 ³ 30) to test a hypothesis about a single proportion, use 

   

   3.     We have a lower-tailed test at the a=.01 level of significance, so z_a =-2.33. Graphically we have:

   

                    The resulting decision rule is:

   If -2.33 £ z then do not reject H₀.

   Otherwise reject H₀.

   4.     Using our sample data (and noting that 5 out of 30 sample bags are overfilled), we calculate the test statistic:

   

   5.    Implementation of our decision rule

   -2.33 £ -1.054 so do not reject H₀

   leads us to conclude that the sample results do not support the conclusion that at least one-quarter of bags of chips are overfilled.

3. Suppose the Quality Control Manager of this potato chip manufacturer also wants to determine if the average weight of all 16 oz. bags of his brand is actually at least 16 ounces. His experience leads him to believe that the weight of the 16 oz. bags is normally distributed. Use the data that he has collected from the thirty recent production runs (referred to in the previous problem) to test this claim at the a=.01 level of significance.

This problem involves testing a hypothesis where our random variable is the ounce weight of a bag of chips. Thus we are dealing with a ratio variable and so will be testing a hypothesis about a single mean. The steps in this hypothesis test are:

1.    H₀:m ³ 16.0 where m is the mean ounce weight of the bags of chips.

       H_a:m < 16.0

2.     Since we are using a large sample (n = 30 ³ 30) to test a hypothesis about a single mean, use

3.    We have a lower-tailed test at the a=.01 level of significance, so z_a=-2.33. Graphically we have:

  

                 The resulting decision rule is:

If -2.33 £ z then do not reject H₀.

Otherwise reject H₀.

4.    Using our sample data we calculate

            and

             which we then use to calculate the test statistic:

5.   Implementation of our decision rule

-2.33 > -3.526 so reject H₀

leads us to conclude that the sample results do not support the conclusion that the mean weight of the bags of chips is at least 16 ounces.

      

4. Wayne John has decided that, in order to diversify his personal investment portfolio, he will invest in a bioengeneering research firm. He decides to evaluate Jones’ Clones, Inc., a relatively small Vermont-based firm that does research on the human genome project. He randomly selects seven months over the past two year period and records the percentage stock price change achieved by Jones’ Clones, Inc., in those months:

   Month & Year	Percentage Change in Stock Price
   May 1999	1.73%
   January 2000	1.37%
   September 2000	1.14%
   April 1999	1.87%
   June 2000	1.30%
   September 1999	1.44%
   December 1998	2.14%

   Provide Mr. John with an appropriate estimate, at the 99% level of confidence, of the monthly change in stock price achieved by Jones’ Clones, Inc. over the past twenty-four months. Comment on the usefulness of your estimate.

   This problem involves testing a hypothesis where our random variable is monthly change in stock price achieved by Jones’ Clones, Inc. Thus we are dealing with an interval variable and so will be constructing a confidence interval for the population mean.

   Using our sample data we calculate

   

               and

   

   which we then use (in conjunction with the value t = ±3.707 that corresponds to a 99% level of confidence and n - 1 = 6 degrees of freedom) to calculate the confidence interval:

   

                I would certainly be concerned over the small sample. However, what would concern me more is the trend in this data. If you look at a runs plot (a scatter plot over time) of the monthly returns, you see

   

               that the percentage change in stock price appears to be dropping over time!

5. An auditor believes that small business who pay their bills late still tend to pay their bills within ten days of their due date. She randomly selects a recently-paid overdue bill from each of fourteen randomly selected small business accounts, then records the number of days each of these bills was outstanding when it was paid. Her sample data are: 

   Account Number	Days Outstanding When Paid
   0356l	22
   04652	12
   02142	19
   17539	20
   08218	23
   29929	24
   10894	14
   00536	8
   01742	16
   120l0	26
   09911	23
   08442	12
   173l9	17
   102l5	33

   The auditor has also indicated the she believes the number of days an overdue bill is outstanding when paid is normally distributed. Use the data that she has collected to test this claim at the a=.05 level of significance.

   This problem involves testing a hypothesis where our random variable is the number of days past due when small businesses pay their late bills. Thus we are dealing with a ratio variable and so will be testing a hypothesis about a single mean. The steps in this hypothesis test are:

   1.   H₀:m £ 10.0 where m is the mean number of days past due when late accounts are paid.

         H_a:m > 10.0

   2.   Since we are using a small sample (n = 14 < 30) taken from a normal population to test a hypothesis about a single mean, use

   

   3.     We have an upper-tailed test at the a=.05 level of significance with 13 degrees of freedom, so t_a=1.771. Graphically we have:

   

            The resulting decision rule is:

   If t_a £ 1.771 then do not reject H₀.

   Otherwise reject H₀.

   4.    Using our sample data we calculate

   

           and

   

            which we then use to calculate the test statistic:

   

   5.    Implementation of our decision rule

   5.2142 > 1.771 so reject H₀

   leads us to conclude that the sample results do not support the conclusion that small business who pay their bills late still tend to pay their bills within ten days of their due date.

         

6. AT&T believes their market share among single consumers has fallen below 50%. To test this theory they randomly telephone thirty single people and ask them to which long-distance telephone service they subscribe. The results are

   Respondent	Response	Respondent	Response
   01	AT&T	16	AT&T
   02	MCI	17	AT&T
   03	AT&T	18	MCI
   04	AT&T	19	Sprint
   05	AT&T	20	AT&T
   06	Sprint	21	Sprint
   07	AT&T	22	AT&T
   08	MCI	23	MCI
   09	AT&T	24	MCI
   10	AT&T	25	AT&T
   11	MCI	26	AT&T
   12	Sprint	27	MCI
   13	AT&T	28	AT&T
   14	AT&T	29	AT&T
   15	Sprint	30	MCI

    Use these sample results to test the appropriate hypothesis at an a=.05 level of significance.

   This problem involves testing a hypothesis where our random variable is whether a single (non-married) consumer uses AT&T long distance service. Thus we are dealing with a nominal variable and so will be testing a hypothesis about a single proportion. The steps in this hypothesis test are:

   1.  H₀:p ³ 0.50 where p is the proportion of bags of chips that are overfilled.

        H_a:p < 0.50

   2.  Since we are using a large sample (np₀ = 30(0.50) =15 ³ 30 & n(1-p₀) = 30(1-0.50) = 15 ³ 30)  to test a hypothesis about a single proportion, use 

   

   3.   We have an upper-tailed test at the a=.05 level of significance, so z_a=1.65. Graphically we have:

   

                    The resulting decision rule is:

   If -1.65 < z then do not reject H₀.

   Otherwise reject H₀.

   4.   Using our sample data (and noting that 17 out of 30 sample bags are overfilled), we calculate the test statistic:

   

   5.  Implementation of our decision rule

   -1.65 £ 0.7303 so do not reject H₀

   leads us to conclude that the sample results do not support the conclusion AT&T's share of the long distance market has fallen below 50%

7. Based on the sample results in the previous problem, provide AT&T with an estimate of Sprint’s market share among single consumers at the 90% level of confidence.

This problem involves testing a hypothesis where our random variable is whether a single (non-married) consumer uses Sprint long distance service. Thus we are dealing with a nominal variable and so will be constructing a confidence interval for the population proportion.

Using our sample data (and recognizing that 5 of our 30 sample respondents use Sprint long distance service) we have

Our sample is so small we end up with a rather imprecise interval! I'm not sure that such an estimate really provides AT&T with any insight.

8. A recent television advertisement for Excedrin^Ó brand aspirin claims that ‘clinical tests prove that Excedrin^Ó is effective in relieving the pain of migraine headaches.’ Use what you have learned in our classroom discussions and assigned readings (as well as your common sense) to develop a one-paragraph critical evaluation of this claim.

Clinical tests involve sample data, which can only suggest results. The only way to conclusively 'prove' a claim would be to take a census, and how do you take a census of all headaches. The advertisement badly misspeaks (and misrepresents the results of the clinical trials) when it  claims that ‘clinical tests prove that Excedrin^Ó is effective in relieving the pain of migraine headaches.’

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