31.A new weight-watching company, Weight Reducers International, advertises that those who join will lose, on the average, 10 pounds the first two weeks with a standard deviation of 2.8 pounds. A random sample of 50 people who joined the new weight reduction program revealed the mean loss to be 9 pounds. At the .05 level of significance, can we conclude that those joining Weight Reducers on average will lose less than 10 pounds? Determine the p-value.
Solution:
Null Hypothesis: H0: m0 = 10 vs. Ha:ma 10
Rejection region can be defined as
z < -za
Here significance level is 0.05,
So, z0.05 = 1.676551 [critical value](Using Statistical Ratio Calculator
From for calculating z with 0.05 significance with df=49)
Now,
z = (X - μ) / σx
Where X is a normal random variable, μ is the mean, and σ is the standard deviation.
Where n is the sample size.
Calculating t-test statistics:
z = (9-10)/(2.8/√50)=-2.525
Calculating p-value:
Degree of freedom = DF = 50-1 = 49
P(z49 -2.525) = 0.007428 (Using http://www.danielsoper.com/statcalc/calc08.aspx
with z = -2.525 and DF = 49)
The p-value of 0.74% is less than significant level of 5% and -2.525 is less than . Hence, we can reject the null hypothesis and conclude that those joining Weight Reducers on average will lose less than 10 pounds.
32. Dole Pineapple, Inc., is concerned that the 16-ounce can of sliced pineapple is being overfilled. Assume the standard deviation of the process is .03 ounces. The quality control department took a random sample of 50 cans and found that the arithmetic mean weight was 16.05 ounces. At the 5 percent level of significance, can we conclude that the mean weight is greater than 16 ounces? Determine the p-value.
Solution:
Null Hypothesis: H0: m0 <= 16 vs. Ha:ma 16
Rejection region can be defined as
z > za
Here significance level is 0.05,
So, z0.05 = 1.676551 [critical value](Using Statistical Ratio Calculator
From http://www.danielsoper.com/statcalc/calc10.aspx for calculating z with 0.05 significance with df=49)
Now,
z = (X - μ) / σx
Where X is a normal random variable, μ is the mean, and σ is the standard deviation.
Where n is the sample size.
Calculating t-test statistics:
z = (16.05-16)/(0.03/√ 50) = 11.785
Calculating p-value:
Degree of freedom = DF = 50-1 = 49
P(z49 11.785) = 0.000000 (Using http://www.danielsoper.com/statcalc/calc08.aspx
with t = 11.785and DF = 49)
The p-value of 0.00% is less than significant level of 5% and calculated statistics 11.785 is greater than critical value 1.645. Hence, we can reject the null hypothesis and conclude that the mean weight is greater than 16 ounces.
38. A recent article in The Wall Street Journal reported that the 30-year mortgage rate is now less than 6 percent. A sample of eight small banks in the Midwest revealed the following 30-year rates (in percent):
4.8 5.3 6.5 4.8 6.1 5.8 6.2 5.6
At the .01 significance level, can we conclude that the 30-year mortgage rate for small banks is less than 6 percent? Estimate the p-value.
Solution:
Mean of the sample = X = 5.6375
Standard deviation of the sample = σ = 0.63457
Null Hypothesis: H0: m0 = 6 vs. Ha:ma 6
Rejection region can be defined as
t < -ta
Here significance level is 0.01,
So, t0.01 = - 2.998 [critical value](Using Statistical Ratio Calculator
From http://www.graphpad.com/quickcalcs/DistMenu.cfm for calculating z with 0.01 significance)
Now,
t = (X - μ) / σx
Where X is a normal random variable, μ is the mean, and σ is the standard deviation.
Where n is the sample size.
Calculating t-test statistics:
t = (5.6375-6)/(0.63457/√ 8) = -1.616
Calculating p-value:
Degree of freedom = DF = 8-1 = 7
P(t8 -1.616) = 0.075064 (Using http://www.danielsoper.com/statcalc/calc08.aspx
with t = -1.616 and DF = 7)
The p-value is 7.5% which is greater than significance level of 1%. Hence, we fail to reject the null hypothesis but we can’t conclude that the rates are less than 6%.
27. A recent study focused on the number of times men and women who live alone buy take-out dinner in a month. The information is summarized below.
Statistic MenWomen
Sample mean 24.5122.69
Population standard deviation 4.48 3.86
Sample size&n bsp;3540
At the .01 significance level, is there a difference in the mean number of times men and women order take-out dinners in a month? What is the p-value?
Solution:
We can define two-tailed statistics for above observations as follows:
Null Hypothesis: H0: (m1 - m2) =0 vs. Ha: (m1 - m2) ¹0
Formula for test statistics,
Using above two-tailed statistics,
Rejection region can be defined as
z < -za/2 or z > za/2
Here significance level is 0.01,
So, z0.005 = 2.5758 (Using Statistical Ratio Calculator
From http://www.graphpad.com/quickcalcs/DistMenu.cfm for calculating z with 0.01 significance)
I am assuming the sample is selected independently and randomly from population. Population size is sufficiently large in both samples.
Calculating test statistics for both samples,
z =
Calculating p-value:
P(|z| 1.871) = 0.0654 (Using http://www.graphpad.com/quickcalcs/Pvalue2.cfm
with z = 1.871 and DF=35+40-2=73)
Since calculated test-statistics is equal to 1.871 which is less than 2.5758. Hence, we cannot reject the null hypothesis as there is not enough evidence at 1% level of significance that there is a difference in the mean number of times men and women order take-out dinners in a month.
46. Grand Strand Family Medical Center is specifically set up to treat minor medical emergencies for visitors to the Myrtle Beach area. There are two facilities, one in the Little River Area and the other in Murrells Inlet. The Quality Assurance Department wishes to compare the mean waiting time for patients at the two locations. Samples of the waiting times, reported in minutes, follow:
Location& nbsp;Waiting Time
Little River31.73 28.77 29.53 22.08 29.47 18.60 32.94 25.18 29.82 26.49
Murrells Inlet 22.93 23.92 26.92 27.20 26.44 25.62 30.61 29.44 23.09 23.10 26.69 22.31
Assume the population standard deviation are not the same. At the .05 significance level, is there a difference in mean waiting time?
Solution:
Size of sample for Little River = 10
Mean of the sample for Little River = X = 27.46
Standard deviation of the sample for Little River = σ = 4.44
Size of sample for Murrells Inlet= 12
Mean of the sample for Murrells Inlet= X = 25.69
Standard deviation of the sample for Murrells Inlet= σ = 2.68
We can define two-tailed statistics for above observations as follows:
Null Hypothesis: H0: (m1 - m2) =0 vs. Ha: (m1 - m2) ¹0
Formula for test statistics,
Using above two-tailed statistics,
Rejection region can be defined as
z < -za/2 or z > za/2
Here significance level is 0.05,
So, z0.025 = 2.0860 (Using Statistical Ratio Calculator
From http://www.graphpad.com/quickcalcs/DistMenu.cfm for calculating z with 0.05 significance and DF = 10+12-2 = 20)
I am assuming the sample is selected independently and randomly from population. Population size is sufficiently large in both samples.
Calculating test statistics for both samples,
z =
Conclusion: Since calculated test-statistics is equal to 1.104 which is less than 2.0860. Hence, we cannot reject the null hypothesis as there is not enough evidence at 5% level of significance that there is a difference in mean waiting time.
52. The president of the American Insurance Institute wants to compare the yearly costs of auto insurance offered by two leading companies. He selects a sample of 15 families, some with only a single insured driver, others with several teenage drivers, and pays each family a stipend to contact the two companies and ask for a price quote. To make the data comparable, certain features, such as the deductible amount and limits of liability, are standardized. The sample information is reported below. At the .10 significance level, can we conclude that there is a difference in the amounts quoted?
Family / Progressive Car Insurance / GEICO Mutual Insurance /Becker / $2,090 / $1,610
Berry / 1,683 / 1,247
Cobb / 1,402 / 2,327
Debuck / 1,830 / 1,367
DuBrul / 930 / 1,461
Eckroate / 697 / 1,789
German / 1,741 / 1,621
Glasson / 1,129 / 1,914
King / 1,018 / 1,956
Kucic / 1,881 / 1,772
Meredith / 1,571 / 1,375
Obeid / 874 / 1,527
Price / 1,579 / 1,767
Phillips / 1,577 / 1,636
Tresize / 860 / 1,188
Solution:
Statistics Progressive Car Insurance GEICO Mutual Insurance
Mean 1390.80 1637.13
SD 437.76 299.06
SEM 113.03 77.22
N 15 15
We can define two-tailed statistics for above observations as follows:
Null Hypothesis: H0: (m1 - m2) =0 vs. Ha: (m1 - m2) ¹0
Formula for test statistics,
Using above two-tailed statistics,
Rejection region can be defined as
z < -za/2 or z > za/2
Here significance level is 0.10,
So, z0.05 = ±1.7011 (Using Statistical Ratio Calculator
From http://www.graphpad.com/quickcalcs/DistMenu.cfm for calculating z with 0.10 significance and df = 28)
I am assuming the sample is selected independently and randomly from population. Population size is sufficiently large in both samples.
Calculating test statistics for both samples,
z =
Since calculated test-statistics is equal to -1.7995 which is less than -1.7011. Hence, we can reject the null hypothesis as there is enough evidence at 10% level of significance that there is a difference in the amounts quoted. So, we can conclude that there is difference in the amounts quoted.
23. A real estate agent in the coastal area of Georgia wants to compare the variation in the
selling price of homes on the oceanfront with those one to three blocks from the ocean.
A sample of 21 oceanfront homes sold within the last year revealed the standard deviation
of the selling prices was $45,600. A sample of 18 homes, also sold within the last
year, that were one to three blocks from the ocean revealed that the standard deviation
was $21,330. At the .01 significance level, can we conclude that there is more variation
in the selling prices of the oceanfront homes?
Solution:
Null Hypothesis: H0: (σ1-σ2) = 0 vs. Ha: (σ1-σ2) ¹0
Here significance level is 0.01,
So, F0.005 = 3.6073 (Using Statistical Ratio Calculator
From http://www.graphpad.com/quickcalcs/DistMenu.cfm for calculating F with 0.01 significance and DF1 = 21-1 = 20 and DF2 = 18-1 = 17)
Calculating test statistics for both samples,
F = (s1/ s2)2 = (45600/21330)2 = 4.5703
Since calculated test-statistics is equal to 4.5703which is greater than 3.1615. Hence, we reject the null hypothesis at 1% level of significance that there is a variation in the selling prices of the oceanfront homes.
28. The following is a partial ANOVA table.
;Sum of ; Mean
Source SquaresdfSquare F
Treatment&nb sp;2
Error&nbs p;20
Total ;50011
Complete the table and answer the following questions. Use the .05 significance level.
Treatment / 2 / 320 / 160 / 8
Error / 9 / 180 / 20
Variation / 11 / 500 / 45.45
a. How many treatments are there?
Answer. 2 + 1 = 3
b. What is the total sample size?
Answer. 11 + 1 = 12
c. What is the critical value of F?
DF for error = 12-3 = 9
Sum of squares for error = 20*9 = 180
Sum of squares for treatment = 500-180 = 320
F = 160/20 = 8
Here significance level is 0.05, Critical value
So, F0.05 = 4.2565
(Using Statistical Ratio Calculator
From http://www.graphpad.com/quickcalcs/DistMenu.cfm for calculating F with 0.05 significance and DF1 = 2 and DF2 = 9)
d. Write out the null and alternate hypotheses.
Null Hypothesis: H0: (m1 - m2) =0 vs. Ha: (m1 - m2) ¹0
e. What is your conclusion regarding the null hypothesis?
Conclusion: Since calculated test-statistics is equal to 8 which is greater than 4.2565. Hence, we reject the null hypothesis as there is sufficient evidence at 5% level of significance that there is a difference in group means.
19. In a particular market there are three commercial television stations, each with its own evening news program from 6:00 to 6:30 P.M. According to a report in this morning’s local newspaper, a random sample of 150 viewers last night revealed 53 watched the news on WNAE (channel 5), 64 watched on WRRN (channel 11), and 33 on WSPD (channel 13). At the .05 significance level, is there a difference in the proportion of viewers watching the three channels?
Null Hypothesis: H0: p1 = p2 = p3 vs. Ha: p1 != p2 != p3
Rejection Region:
Calculate chi-square statistics > Critical chi-square statistics
We are going to use chi-square test.
If p1 = p2 = p3, then p1 = p2 = p3 = 150/3 = 50
Here significance level is 0.05, Critical value
So, Chi square = 5.9915
(Using Statistical Ratio Calculator
From http://www.graphpad.com/quickcalcs/DistMenu.cfm for calculating Chi square with 0.05 significance and DF = 3-1 = 2)
Calculate chi-square statistics,
Chi square = (53-50)2/50 + (64-50)2/50 +(33-50)2/50 = 9.88
Conclusion: Since calculated Chi square -statistics is equal to 9.88 which is greater than 5.9915. Hence, we reject the null hypothesis as there is sufficient evidence at 5% level of significance that there is a difference in the proportion of viewers watching the three channels.
20. There are four entrances to the Government Center Building in downtown Philadelphia. The building maintenance supervisor would like to know if the entrances are equally utilized. To investigate, 400 people were observed entering the building. The number using each entrance is reported below. At the .01 significance level, is there a difference in the use of the four entrances?
EntranceFrequency
Main Street 140
Broad Street120
Cherry Street90
Walnut Street50
Total400
Solution:
Null Hypothesis: H0: E1 = E2 = E3 = E4 vs. Ha: E1 != E2 != E3 != E4
Rejection Region:
Calculate chi-square statistics > Critical chi-square statistics
Here significance level is 0.01, Critical value
So, Chi square = 11.3449
(Using Statistical Ratio Calculator
From http://www.graphpad.com/quickcalcs/DistMenu.cfm for calculating Chi square with 0.01 significance and DF = 4-1 = 3)
Calculate chi-square statistics,
Chi square = (140-100)2/100 + (120-100)2/100 +(90-100)2/100 +(50-100)2/100 = 46.00
Conclusion: Since calculated Chi square -statistics is equal to 46.00 which is greater than 11.3449. Hence, we reject the null hypothesis as there is sufficient evidence at 1% level of significance that there is a difference in the use of the four entrances.