Lind Chapter 9: Exercises 12, 28
12. The American Sugar Producers Association wants to estimate the mean yearly sugar consumption. A sample of 16 people reveals the mean yearly consumption to be 60 pounds with a standard deviation of 20 pounds.
a. What is the value of the population mean? What is the best estimate of this value?
Population Mean = 60 pounds. The best estimate for the population mean is the sample mean.
b. Explain why we need to use the t distribution. What assumption do you need to make?
Since we have a small sample size (Less than 30), then we need to use the Student’s t Distribution. We need to make the assumption that the sample is normally distributed.
c. For a 90 percent confidence interval, what is the value of t?
For a 90% confidence interval and df = n – 1 = 16 – 1 = 15, the value of t = 1.753
d. Develop the 90 percent confidence interval for the population mean.
=60 – 8.765 < µ < 60 + 8.765
= 51.24 < µ < 68.77
e. Would it be reasonable to conclude that the population mean is 63 pounds?
Yes, since the confidence interval includes 63, we can conclude, with 90% confidence, that the population mean is 63 pounds.
28. A processor of carrots cuts the green top off each carrot, washes the carrots, and inserts six to a package. Twenty packages are inserted in a box for shipment. To test the weight of the boxes, a few were checked. The mean weight was 20.4 pounds, the standard deviation 0.5 pounds. How many boxes must the processor sample to be 95 percent confident that the sample mean does not differ from the population mean by more than 0.2 pounds?
Sample size =
Since E = 0.2, = 0.5, and Z0.95 = 1.96
n =
= 24.01, rounded up to 25 packages
Lind Chapter 10: Exercises 6, 18
6. The MacBurger restaurant chain claims that the waiting time of customers for service is normally distributed, with a mean of 3 minutes and a standard deviation of 1 minute. The quality-assurance department found in a sample of 50 customers at the Warren Road MacBurger that the mean waiting time was 2.75 minutes. At the .05 significance level, can we conclude that the mean waiting time is less than 3 minutes?
A) Hypothesis:
B): Level of Significance:
C) Test Statistic
=
= -1.77
D) Decision Rule:
Reject H0 if the computed test statistic is less than -1.645
E) Conclusion:
At the 5% level of significance, the data provide sufficient evidence to conclude that the mean waiting time is less than 3 minutes.
18. The management of White Industries is considering a new method of assembling its golf cart. The present method requires 42.3 minutes, on the average, to assemble a cart. The mean assembly time for a random sample of 24 carts, using the new method, was 40.6 minutes, and the standard deviation of the sample was 2.7 minutes. Using the .10 level of significance, can we conclude that the assembly time using the new method is faster?
A)Hypothesis:
B)Level of Significance:
C)Test Statistic:
=
= -3.08
D)Decision Rule:
For = 0.10, n = 24, df = n-1 = 24-1 = 23 (One-tailed test):
Reject H0 if t < -1.319
E)Conclusion:
Reject the null hypothesis. At the 0.1 level of significance, the data provides sufficient evidence to reject H0, and conclude that the assembly time using the new method is faster
Lind Chapter 11: Exercises 4, 16, 42
4. As part of a study of corporate employees, the Director of Human Resources for PNC, Inc. wants to compare the distance traveled to work by employees at their office in downtown Cincinnati with the distance for those in downtown Pittsburgh. A sample of 35 Cincinnati employees showed they travel a mean of 370 miles per month, with a standard deviation of 30 miles per month. A sample of 40 Pittsburgh employees showed they travel a mean of 380 miles per month, with a standard deviation of 26 miles per month. At the .05 significance level, is there a difference in the mean number of miles traveled per month between Cincinnati and Pittsburgh employees? Use the five-step hypothesis-testing procedure.
Step 1:Ho: c = pH1: cp
Step 2:The 0.05 significance level was chosen
Step 3:Reject Ho and accept H1 if z less than 1.96 or greater than 1.96
Step 4:1.53, found by
Step 5:Fail to reject Ho. There is no difference in the mean number of miles traveled per month. The p-value is 0.126 found by 2(0.5000 – 0.4370)
16)
Ho: sdH1: sddf = 15 + 12 – 2 = 25Reject Ho if t > 2.485
Do not reject Ho. There is no difference in the mean amount of time spent watching television.
42)
Ho: d = 0H1: d 0Reject Ho if z < 1.761 or z > 1.761
= 246sd = 547
Do not reject Ho. There is no difference in the mean insurance price.
Lind Chapter 15: Exercises 12, and 26
12)
Day / Rounds(Observed) / Rounds
(Expected) / O-E / (O-E)^2 / (O-E)^2/E
ABC / 165 / 150 / 15 / 225 / 1.5
CBS / 140 / 150 / -10 / 100 / 0.666667
NBC / 125 / 150 / -25 / 625 / 4.166667
Cable / 70 / 50 / 20 / 400 / 8
Total / 500 / 500 / 0 / 1350 / 14.33
1)Hypothesis:
H0: The guideline is still reasonable
H1: The guideline is not still reasonable
2)Level of Significance:
= 0.05
3)Test Statistic:
= 14.33 (Calculations are on the attached excel sheet)
4)Critical Region:
For = 0.05, with df = k -1 = 4 -1 = 3, we reject H0 for any test statistic greater than 7.815
5)
Since the computed test statistic = 14.33 > 7.815, we reject H0. Therefore at the 5% level of significance, we conclude that the guideline is not still reasonable.
26)
Degree of Job PressureAge
(Years) / Low / Medium / High
Less than 25 / 20 / 18 / 22
25 up to 40 / 50 / 46 / 44
40 up to 60 / 58 / 63 / 59
60 and older / 34 / 43 / 43
A)Hypothesis:
H0: There is no relationship between job pressure and age
H1: There is a relationship between job pressure and age
2) Level of Significance:
=0.01
C) Test Statistic:
= 2.191
D) Decision Rule:
For =0.01
df = (R-1)(C-1) = (4-1)(3-1) = 3*2 = 6
Reject H0 if computed test statistic > 16.81.
E) Conclusion:
Since the computed test statistic = 2.191 < 16.81, then we fail to reject H0, and conclude that there is no relationship between job pressure and age. This is because the Chi Square test of independence tests whether 2 variables are dependent or not. Hence, accepting the null hypothesis- as indicated in step 1, would mean that the variables are not related, i.e. independent.
Lind Chapter 12: Exercises 10, 18, and 26
10)
Banking / Retail / Insurance12 / 8 / 10
10 / 8 / 8
10 / 6 / 6
12 / 8 / 8
10 / 10 / 10
Hypothesis:
Level of Significance:
Decision Rule:
If F > 3.885299, reject H0
ANOVA: Single FactorSUMMARY
Groups / Count / Sum / Average / Variance
Banking / 5 / 54 / 10.8 / 1.2
Retail / 5 / 40 / 8 / 2
Insurance / 5 / 42 / 8.4 / 2.8
ANOVA
Source of Variation / SS / df / MS / F / P-value / F crit
Between Groups / 22.93 / 2 / 11.46667 / 5.733333 / 0.01788 / 3.88529
Within Groups / 24 / 12 / 2
Total / 46.93 / 14
Since the calculated Test Statistic F (5.7333) is greater than the critical value (3.88529), we reject H0. This means that there is a difference in the mean number of hours spent at a terminal per week by industry.
18)
ANOVA: Two-Factor Without ReplicationSUMMARY / Count / Sum / Average / Variance
Monday / 3 / 56 / 18.66667 / 25.33333
Tuesday / 3 / 58 / 19.33333 / 25.33333
Wednesday / 3 / 52 / 17.33333 / 17.33333
Thursday / 3 / 60 / 20 / 4
Friday / 3 / 72 / 24 / 16
St Luke's / 5 / 88 / 17.6 / 6.8
St Vincent / 5 / 112 / 22.4 / 14.8
Mercy / 5 / 98 / 19.6 / 26.8
ANOVA
Source of Variation / SS / df / MS / F / P-value / F crit
Days / 75.73333 / 4 / 18.93333 / 1.285068 / 0.352359 / 3.837853
Hospital / 58.13333 / 2 / 29.06667 / 1.972851 / 0.201147 / 4.45897
Error / 117.8667 / 8 / 14.73333
Total / 251.7333 / 14
As both the p-values for different days of a week and different hospitals are greater than 0.05, we can say that there are no significant differences in the mean number of surgeries performed by hospital or by day of the week for alpha= 0.05.
26)
Ho: 1 = 2 = 3H1: Not all means are equal
Ho is rejected if F > 3.89
SourceSSdfMSF
Treatment26.13213.06713.52
Error11.60120.967
Total37.7314
Ho is rejected since 13.52 > 3.89. There is a difference in the mean weight loss among the three
diets.