Question

A typical washing machine has several different cycles, including soak, wash and rinse. The emergency consumption of the washing machine is linked to the length of each cycle. A random sample of 21 washing machines was selected and the length (in minutes) of each wash cycle was recorded. The resulting sample mean and standard deviation are respectively 37.8 and 5.9.

a) Find the value of the point estimate of the mean wash cycle time of this type of washing machine.

b) Find the value of the point estimate of the standard deviation of the wash cycle time of this type of washing machine.

c) If the wash cycle time of this type of washing machine is approximately normally distributed, construct a 95% confidence interval for the true mean wash cycle time of this type of washing machine.

d) Interpret the resulting confidence interval in part (c).

e) Suppose the confidence interval obtained in part (c) is too wide. How can the width of this interval be reduced? Discuss all possible alternatives. Which alternative is the best?

f) Using the provided statistical evidence, can we conclude that the mean wash cycle time of this type of washing machine is different from 41.5? Use the critical value approach. State the required assumption(s) for this test of hypothesis.

Answer 1

a)

point estimate of the mean wash cycle time of this type of washing machine = 37.8

b)

point estimate of the standard deviation of the wash cycle time of this type of washing machine = 5.9

c)

Level of Significance ,    α =    0.05          
degree of freedom=   DF=n-1=   20          
't value='   tα/2=   2.0860   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   5.9 / √   21   =   1.287
margin of error , E=t*SE =   2.0860   *   1.28749   =   2.686
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    37.80   -   2.686   =   35.114
Interval Upper Limit = x̅ + E =    37.80   +   2.686   =   40.486
95%   confidence interval is (   35.1   < µ <   40.5   )

d)

CI is ( 35.1   < µ <   40.5) it implies that true mean wash cycle time of this type of washing machine would fall in between this range

e)

width of this interval cab be reduced :

-by increasing level of significance(alpha value)

-by increasing sample size

f)

Ho :   µ =   41.5                  
Ha :   µ ╪   41.5       (Two tail test)          
                          
Level of Significance ,    α =    0.05                  
sample std dev ,    s =    5.9000                  
Sample Size ,   n =    21                  
Sample Mean,    x̅ =   37.8000                  
                          
degree of freedom=   DF=n-1=   20                  
                          
Standard Error , SE = s/√n =   5.9000   / √    21   =   1.2875      
t-test statistic= (x̅ - µ )/SE = (   37.800   -   41.5   ) /    1.2875   =   -2.87
                          
critical t value, t* =       2.0860   [Excel formula =t.inv(α/no. of tails,df) ]         

decision: |test stat| > critical value , reject Ho

conclusion: there is enough evidence to conclude that  mean wash cycle time of this type of washing machine is different from 41.5