Question

A) The mean life of a battery used in a digital clock is 305 days. The lives of the batteries follow the normal distribution. The battery was recently modified to last longer. A sample of 20 of the modified batteries had a mean life of 311 days with a sample standard deviation of 12 days. Did the modification increase the mean life of the battery? State the null hypothesis and the alternate hypothesis. Show the decision rule graphically. Use the .05 significance level. What is your decision regarding the null hypothesis? Briefly summarize your results.

B) Anonymous manufactures and assembles desks and other office equipment at several plants in California. The weekly production of the Model PCC R206 desk at the Pasadena Plant follows a normal probability distribution with a mean of 200 and a population standard deviation of 16. Recently, because of market expansion, new production methods have been introduced and new employees hired. The vice president of manufacturing would like to investigate whether there has been a changein the weekly production of the Model PCC R206 desk. Is the mean number of desks produced at the Pasadena Plant different from 200 at the 0.01 significance level? The company’s internal sampling using 50 weeks, because the plant was shut down for 2 weeks due to vacation and maintenance showed a mean number of desks at 203.5

C) Heinz, a manufacturer of ketchup, uses a particular machine to dispense 16 ounces of its ketchup into containers. From many years of experience with the particular dispensing machine, Heinz knows the amount of product in each container follows a normal distribution with a mean of 16 ounces and a standard deviation of 0.15 ounce. A sample of 50 containers filled last hour revealed the mean amount per container was 16.015 ounces. Does this evidence suggest that the mean amount dispensed is different from 16 ounces? Use the .05 significance level. State the null hypothesis and the alternate hypothesis. What is the probability of a Type I error? Give the formula for the test statistic. State the decision rule. Determine the value of the test statistic. What is your decision regarding the null hypothesis? Interpret the result.

D) The American Water Works Association reports that the per capita water use in a single-family home is 69 gallons per day. Legacy Ranch is a relatively new housing development consisting of 100 homes. The builders installed more efficient water fixtures, such as low-flush toilets, and subsequently conducted a survey of the residences. Thirty-six homes responded, and the sample mean water use per day was 64 gallons with a population standard deviation of 8.8 gallons per day. At the .10 level of significance, is that enough evidence to conclude that residents of Legacy Ranch use less water on average?

Answer 1

1)

Ho :   µ =   305  
Ha :   µ >   305   (Right tail test)
          
Level of Significance ,    α =    0.050  
sample std dev ,    s =    12.0000  
Sample Size ,   n =    20  
Sample Mean,    x̅ =   311.0000  
          
degree of freedom=   DF=n-1=   19  
          
Standard Error , SE = s/√n =   12/√20=   2.6833  
t-test statistic= (x̅ - µ )/SE =    (311-305)/2.6833=   2.236  
          
          
p-Value   =   0.0188   [Excel formula =t.dist(t-stat,df) ]
Decision:   p-value≤α, Reject null hypothesis       
...............

2)

  
Ho :   µ =   200  
Ha :   µ ╪   200   (Two tail test)
          
Level of Significance ,    α =    0.010  
population std dev ,    σ =    16.0000  
Sample Size ,   n =    50  
Sample Mean,    x̅ =   203.5000  
          
'   '   '  
          
Standard Error , SE = σ/√n =   16/√50=   2.2627  
Z-test statistic= (x̅ - µ )/SE =    (203.5-200)/2.2627=   1.547  

          
p-Value   =   0.1219   [ Excel formula =NORMSDIST(z) ]
Decision:   p-value>α, Do not reject null hypothesis

..............

3)

  Ho :   µ =   16  
Ha :   µ ╪   16   (Two tail test)
          
Level of Significance ,    α =    0.050  
population std dev ,    σ =    0.1500  
Sample Size ,   n =    50  
Sample Mean,    x̅ =   16.0150  
          
'   '   '  
          
Standard Error , SE = σ/√n =   0.15/√50=   0.0212  
Z-test statistic= (x̅ - µ )/SE =    (16.015-16)/0.0212=   0.707  
          
          
p-Value   =   0.4795   [ Excel formula =NORMSDIST(z) ]
Decision:   p-value>α, Do not reject null hypothesis

type II error

.......

4)

Ho :   µ =   69  
Ha :   µ <   69   (Left tail test)
          
Level of Significance ,    α =    0.100  
population std dev ,    σ =    88.0000  
Sample Size ,   n =    36  
Sample Mean,    x̅ =   64.0000  
          
'   '   '  
          
Standard Error , SE = σ/√n =   88/√36=   14.6667  
Z-test statistic= (x̅ - µ )/SE =    (64-69)/14.6667=   -0.341  
          
          
p-Value   =   0.3666   [ Excel formula =NORMSDIST(z) ]
Decision:   p-value>α, Do not reject null hypothesis       
   

............

Please let me know in case of any doubt.

Thanks in advance!

Please upvote!