Math assignment

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One.  Clark Heter is an industrial engineer at Lyons Products. He would like to determine whether there are more units produced on the night shift than on the day shift. A sample of 50 day-shift workers showed that the mean number of units produced was 353, with a population standard deviation of 25. A sample of 55 night-shift workers showed that the mean number of units produced was 363, with a population standard deviation of 31 units.

 

At the .01 significance level, is the number of units produced on the night shift larger?

 

(a) This is a -tailed test.

 

(b) The decision rule is to reject  if Z < . (Negative amount should be indicated by a minus sign. Round your answer to 2 decimal places.)

 

(c) The test statistic is Z = . (Negative amount should be indicated by a minus sign. Round your answer to 2 decimal places.)

 

TWO

Each month the National Association of Purchasing Managers publishes the NAPM index. One of the questions asked on the survey to purchasing agents is: Do you think the economy is contracting? Last month, of the 310 responses, 164 answered yes to the question. This month, 177 of the 291 responses indicated they felt the economy was contracting.

 

At the .02 significance level, can we conclude that a larger proportion of the agents believe the economy is contracting this month?

 

 

pc =  . (Do not round the intermediate value. Round your answer to 2 decimal places.)

 

 

The test statistic is  . (Negative amount should be indicated by a minus sign. Do not round the intermediate value. Round your answer to 2 decimal places.)

 

 

Decision:  the null. H0 : π1 ≥ π2

 

THREE

The manufacturer of an MP3 player wanted to know whether a 10 percent reduction in price is enough to increase the sales of its product. To investigate, the owner randomly selected eight outlets and sold the MP3 player at the reduced price. At seven randomly selected outlets, the MP3 player was sold at the regular price. Reported below is the number of units sold last month at the sampled outlets.

 

  Regular price 133 124 88 112 144 128 96  
  Reduced price  124 134 152 134 114 109 113 114

 

 

At the .050 significance level, can the manufacturer conclude that the price reduction resulted in an increase in sales? Hint: For the calculations, assume the Reduced price as the first sample.

 

 

The pooled variance is . (Round your answer to 2 decimal places.)

 

 

The test statistic is . (Round your answer to 2 decimal places.)

 

 

 H0.

 

 

FOUR

One of the music industry’s most pressing questions is: Can paid download stores contend nose-to-nose with free peer-to-peer download services? Data gathered over the last 12 months show Apple’s iTunes was used by an average of 1.81 million households with a sample standard deviation of .47 million family units. Over the same 12 months WinMX (a no-cost P2P download service) was used by an average of 2.21 million families with a sample standard deviation of .32 million. Assume the population standard deviations are not the same.

 

(A) Find the degrees of freedom for unequal variance test. (Round down your answer to nearest whole number.)

 

 

  Degrees of freedom  

 

 

(B) State the decision rule for .02 significance level: H0: A = W; H1: A ≠ W . (Negative amounts should be indicated by a minus sign. Round your answer to 3 decimal places.)

 

 

  Reject H0 if t <  or t >

 

 

(C) Compute the value of the test statistic. (Negative amount should be indicated by a minus sign.Round your answer to 2 decimal places.)

 

 

  Value of the test statistic  

 

 

(D) Test the hypothesis of no difference in the mean number of households picking either variety of service to download songs. Use the .02 significance level.

 

 

   H0. There is  difference in the mean number of households picking either variety of service to download songs.  

 

 

 

 

FIVE

When only two treatments are involved, ANOVA and the Student t test (Chapter 11) result in the same conclusions. Also, . As an example, suppose that 14 randomly selected students were divided into two groups, one consisting of 6 students and the other of 8. One group was taught using a combination of lecture and programmed instruction, the other using a combination of lecture and television. At the end of the course, each group was given a 50-item test. The following is a list of the number correct for each of the two groups. Using analysis of variance techniques, test the null hypothesis,  that the two mean test scores are equal.

 

Lecture and Programmed Instruction Lecture and Television
12 37
14 27
26 37
28 21
13 25
15 27
  26
  23
 

 

 

(a-1) Complete the ANOVA table. (Round SS, MS and F values to 2 decimal places.)

 

 

  Source SS df MS F
  Factors        
  Error        
         
  Total        
   

 

  

 

    
 

 

 

(a-2) Use a  level of significance. (Round your answer to 2 decimal places.)

 

 

  The test statistic is F  

 

 

(b) Using the t test from Chapter 11, compute t. (Negative amount should be indicated by a minus sign. Round your answer to 2 decimal places.)

 

 

  t  

 

 

(c) There is  in the mean test scores

 

 

SIX

The following hypotheses are given.
 
Ho : σ1² ≤ σ2²
H1 : σ1² > σ2²

 

A random sample of five observations from the first population resulted in a standard deviation of 12. A random sample of seven observations from the second population showed a standard deviation of 7. At the .01 significance level, is there more variation in the first population?

 

 

The test statistic is . (Round your answer to 2 decimal places.)
 
Decision:   Ho

 

 

SEVEN

Given the following sample information, test the hypothesis that the treatment means are equal at the .05 significance level.

 

Treatment 1 Treatment 2 Treatment 3
8 3 3
11 2 4
10 1 5
  3 4
  2  

 

 

(a-1) State the null hypothesis and the alternate hypothesis.

 

 

      Null hypothesis

 

 

  Ho: μ1 = μ2 = μ3
  Ho: μ1 = μ2

 

 

a-2) Alternative hypothesis

 

 

  H1: Treatment means are all the same
  H1: Treatment means are not all the same

 

 

b) What is the decision rule? (Round your answer to 2 decimal places.)

 

 

  Reject Ho if F >

 

 

c) Compute SST, SSE, and SS total. (Round your answers to 2 decimal places.)

 

 

   
  SST  
  SSE  
   
  SS total  
   

 
 

 

 

d) Complete an ANOVA table. (Round F, SS to 2 decimal places and MS to 3 decimal places.)

 

 

  Source SS df MS F
  Treatments        
  Error        
         
  Total        

 

 

 

e) State your decision regarding the null hypothesis.

 

 

  Reject H0.
  Do not reject H0.

 

 

f) If H0 is rejected, can we conclude that treatment 1 and treatment 2 differ? Use the 95 percent level of confidence.

 

 

  , we  conclude that the treatments 1 and 2 have different means.

 

 

EIGHT

There are four radio stations in Midland. The stations have different formats (hard rock, classical, country/western, and easy listening), but each is concerned with the number of minutes of music played per hour. From a sample of 10 hours from each station, the following sample means were offered.

 

formula340.mml       formula341.mml     formula342.mml            formula343.mml

 

 

                                                      SS total = 650.75

 

(a) SST = . (Round your answer to 3 decimal places.)

 

 

(b) SSE = . (Round your answer to 3 decimal places.)

 

 

(c) Complete an ANOVA table. (Round SS, MS, F to 3 decimal places and df to nearest whole number.)

 

 

  SS df MS F
Treatments        
Error        
         
Total        
   

 

  

 

    
 

 

 

(d) At the .05 significance level, is there a difference in the treatment means?

 

 

NINE

Given the following sample information, test the hypothesis that the treatment means are equal at the .05 significance level.

 

Treatment 1 Treatment 2 Treatment 3
3 9 6
2 6 3
5 5 5
1 6 5
3 8 5
1 5 4
  4 1
  7 5
  6  
  4  
 

 

 

PictureClick here for the Excel Data File

 

 

(a) Ho : μ1  μ2   μ3.
   
  H1 : Treatment means  all the same

 

 

(b) Reject   Ho  if F >  .(Round your answer to 2 decimal places.)

 

 

(c) SST =   SSE =   SS total =  (Round your answers to 2 decimal places.)

 

 

(d) Complete the ANOVA table. (Round SS, MS and F values to 2 decimal places.)

 

 

  Source SS df MS F
  Treatments        
  Error        
         
  Total        
   

 

  

 

    
 

 

 

(e) Decision:  Ho

 

 

(f) Find the 95% confidence interval for the difference between treatment 2 and 3. (Round your answers to 2 decimal places.)
   
  95% confidence interval is:   ±
   
  We can conclude that the treatments 2 and 3 are

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