Question

In: Statistics and Probability

1)         The population standard deviation for waiting times to be seated at a restaurant is known...

1)         The population standard deviation for waiting times to be seated at a restaurant is known to be 10 minutes. An expensive restaurant claims that the average waiting time for dinner is approximately 1 hour, but we suspect that this claim is inflated to make the restaurant appear more exclusive and successful. A random sample of 30 customers yielded a sample average waiting time of 50 minutes.

  1. Is there evidence to say that the restaurant’s claim is too high? Insert the results from StatCrunch here.
  1. State the hypotheses, your decision and conclusion. Include the reason for your decision using the output from above.
  1. The original data (individual waiting times) is not normally distributed. What theorem allows us to do the calculations in part (a)?

2)         A sample of 800 items produced on a new machine showed that 48 of them are defective. The factory will get rid the machine if the data indicates that the proportion of defective items is significantly more than 5%. At a significance level of 10% does the factory get rid of the machine or not?

  1. Insert the results from StatCrunch here.
  1. State the hypotheses, your decision and conclusion. Include the reason for your decision using the results from above.

3)         A psychologist claims that the mean age at which children start walking is 12.5 months. The following data give the age at which 18 randomly selected children started walking.

15        11        13        14        15        12        15        10        16

17        14        16        13        15        15        14        11        13

  1. Test at the 1% level of significance if the mean age at which children start walking is different from 12.5 months. Insert the results from StatCrunch here.
  1. State the hypotheses, state your decision and conclusion. Include the reason for your decision from the results above.

4)         According to a study, 107 of 507 female college students were on a diet at the time of the study.

a) Construct a 99% confidence interval for the true proportion of all female students who were on a diet at the time of this study. Insert the results from StatCrunch here.

b) Interpret this interval.

c) Is it reasonable to think that only 17% of college women are on a diet? Why or why not?

Solutions

Expert Solution

1)

The original data (individual waiting times) is not normally distributed. but central limit theorem allows us to do the calculations in part (a)

Ho :   µ ≥ 60                  
Ha :   µ <   60       (Left tail test)          
                          

  
population std dev ,    σ =    10.0000                  
Sample Size ,   n =    30                  
Sample Mean,    x̅ =   50.0000                  
                          
'   '   '                  
                          
Standard Error , SE = σ/√n =   10.0000   / √    30   =   1.8257      
Z-test statistic= (x̅ - µ )/SE = (   50.000   -   60   ) /    1.8257   =   -5.477
                          

critical z value, z* =       -2.326   [Excel formula =NORMSINV(α/no. of tails) ]              
                          
p-Value   =   0.0000   [ Excel formula =NORMSDIST(z) ]              
Decision:   p-value<α, Reject null hypothesis                       
conclusion: there is enough evidence to say that the restaurant’s claim is too high

2)

Ho :   p =    0.05                  
H1 :   p >   0.05       (Right tail test)          
                          
Level of Significance,   α =    0.10                  
Number of Items of Interest,   x =   48                  
Sample Size,   n =    800                  
                          
Sample Proportion ,    p̂ = x/n =    0.0600                  
                          
Standard Error ,    SE = √( p(1-p)/n ) =    0.0077                  
Z Test Statistic = ( p̂-p)/SE = (   0.0600   -   0.05   ) /   0.0077   =   1.2978
                          
critical z value =        1.282   [Excel function =NORMSINV(α)              
                          
p-Value   =   0.0972   [Excel function =NORMSDIST(-z)              
Decision:   p-value<α , reject null hypothesis                       
There is enough evidence to conclude that the factory should get rid of the machine

  


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