Records on a fleet of trucks reveal that the average life of a set of spark plugs is normally distributed with a population mean of 22,100 miles. A manufacturer of spark plugs claims that its plugs...

Records on a fleet of trucks reveal that the average life of a set of spark plugs is normally distributed with a population mean of 22,100 miles. A manufacturer of spark plugs claims that its plugs have an average life in excess of 22,100 miles. The fleet owner purchased 18 sets and found that the sample average life was 23,400 miles, the sample standard deviation was 1,500 miles and the calculated value of t = 3.677. Based on these findings,
is there enough evidence
to accept the manufacturer's claim at the 0.05 level.(Explain your answer)
Yes __X___
Explain:
Ho: u = 22,100
H1: u

Reject Ho if p-value is greater than 5%.
(23,400-22,100)/[1300/sqrt(18)] = -4.24
----------------------------
Since the p-value is less than 5%, Fail to reject Ho.
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 standard deviation of 12 days. You have been asked to test if the modifications have increased the mean life of the batteries at the .05 significance level. (Use for Q11 & 12 below)
11. State the null & alternative hypothesis:
Null:
m
= 305; alternative:
m
? 305
12. Compute the test statistic and indicate your decision regarding the null hypothesis? Hint: First determine your decision rule.
Reject Ho if p-value is greater than 5%.
(311-305/[12/sqrt(20)] = -4.47
P-value = P(t
----------------------------
Since the p-value is less than 5%, Fail to reject Ho.
13. As a region sales manager you are interested in testing if there is a difference in sales close rates between two of your sales regions. In the west region you contacted 50 sales people and found that they closed a sale at a rate of 40% of the time. The East region had a higher close rate at 50% based on a sample of 80. In the space below write the formula for the test statistic used to evaluate the difference. Second complete (fill in the values) the formula using the information above.
Do not calculate the final value of the test statistic

The following Analysis of Variance table includes information from three treatments (samples) each with six observations. Suggestion: Fill in the missing values in the table to help answer Q14-16.


14.What are the degrees of freedom for the numerator and denominator?
a.3 and 18
b.2 and 17
c.3 and 15
d.2 and 15
15.What is the critical value of
F
at the 5% level of significance?
a.3.29
b.3.68
c.3.59
d.3.20
16.What is the computed value of
F?Show work!
a.7.48
b.7.84
c.8.84
d.8.48
17. While attending a sales meeting you were asked about the relationship between sales calls and actual sales made. Fortunately, you had just completed a study on this topic but you failed to bring the report with you. However, you do recall that the
coefficient of
determination
was calculated to be .49. What can you report about the strength of relationship between sales calls and actual sales made?
18. The following data are results from a study you recently completed. Calculate the correlation coefficient for this data. (Show work or use excel print out)

X Y 4 4 -1.6 -1.8 2.56 3.24 2.88 5 6 -0.6 0.2 0.36 0.04 -0.12 3 5 -2.6 -0.8 6.76 0.64 2.08 6 7 0.4 1.2 0.16 1.44 0.48 10 7 4.4 1.2 19.36 1.44 5.28 29.2 6.8

19. Given the following information what is the regression equation? The value of r is (minus) -0.8908 with a standard deviation for X= 1.73 & Y= 3.34. The Mean of X = 4.875 and the Mean of Y = 10.625.
20. A strong positive correlation of .95 indicates that the independent variable X is the cause for the variation in Y

1. TRUE


2. FALSE