Concept explainers
(a)
Section 1:
To find: The difference of estimates for each subject.
(a)
Section 1:
Answer to Problem 38E
Solution: The table for difference of estimates is as follows,
Jocko |
Other |
Difference |
1410 |
1250 |
160 |
1550 |
1300 |
250 |
1250 |
1250 |
0 |
1300 |
1200 |
100 |
900 |
950 |
-50 |
1520 |
1575 |
-55 |
1750 |
1600 |
150 |
3600 |
3380 |
220 |
2250 |
2125 |
125 |
2840 |
2600 |
240 |
Explanation of Solution
Calculation: To calculate Difference follow the below mentioned steps in Minitab;
Step 1: Enter the variable name as ‘Jocko’ in column C1 and enter the provided data of ‘Jocko’s’. Enter the variable name as ‘Other’ in column C2 and enter the provided data of ‘Other’.
Step 2: Enter variable name as ‘Difference’ in column C3.
Step 3: Go to
Step 4: In the dialog box that appears, select ‘Difference’ in ‘Store result in variable’.
Step 5: Enter 'Jocko-Other' in ‘Expression’ and click ‘OK’.
From Minitab result, the table for weight change is as follows,
Jocko |
Other |
Difference |
1410 |
1250 |
160 |
1550 |
1300 |
250 |
1250 |
1250 |
0 |
1300 |
1200 |
100 |
900 |
950 |
-50 |
1520 |
1575 |
-55 |
1750 |
1600 |
150 |
3600 |
3380 |
220 |
2250 |
2125 |
125 |
2840 |
2600 |
240 |
Section 2:
To find: The
Section 2:
Answer to Problem 38E
Solution: The required mean is 114.0 and the standard deviation is 114.4
Explanation of Solution
Calculation: Follow the below mentioned steps in Minitab,
Step 1: Follow the step 1 to 5 performed in part (a).
Step 2: Go to
Step 3: In dialog box that appears select ‘Difference’ under the field marked as ‘Variables’. Then click on ‘Statistics’.
Step 4: In dialog box that appears select ‘None’ and then ‘Mean’. Finally, click ‘OK’ on both dialog boxes.
From Minitab result, the mean is 114.0 and the standard deviation is 114.4.
(b)
To test: The hypothesis that the estimates of the two garages do not differ significantly.
(b)
Answer to Problem 38E
Solution: The mean estimates for Jocko and Other differ significantly at 5% significance level.
Explanation of Solution
Calculation:
The null hypothesis is the hypothesis of no difference. If
The null hypothesis assumes that there is no significant difference and is formulated as,
And the alternative hypothesis assumes a significant change in the average estimates and is formulated as,
The formula of test statistic for paired t-test is as follows,
In the above formula,
The test statistic is calculated as,
The formula for degrees of freedom is as follows,
And it is calculated as,
To calculate the P-value follow the below mentioned steps in Minitab,
Step 1: Follow the step 1 to 5 performed in part (a).
Step 2: Go to
Step 3: In the dialogue box that appears, select samples in columns.
Step 4: Enter ‘Jocko’ under the field marked as ‘First sample’ and ‘Other’ under the field marked as ‘Second sample’.
Step 5: In the dialogue box that appears, Click on ‘Option’. Enter 95.0 in the field of Confidence interval, 0.0 in ‘Test mean’ and ‘not equal’ in ‘Alternative’.
From Minitab results, the P-value is 0.012.
Conclusion: Since, the P-value is 0.012 which is less than 0.05, the significance level. So the null hypothesis is rejected and it is concluded that the mean estimate for ‘Jocko’ is different from mean estimate for ‘Other’.
(c)
To find: The confidence interval.
(c)
Answer to Problem 38E
Solution: The required confidence interval is
Explanation of Solution
Calculation:
The confidence interval is an interval for which there is 95% chances that it contains the population parameter (population mean).
To calculate the confidence interval, follow the below mentioned steps in Minitab,
Step 1: Follow the step 1 to 5 performed in part (a).
Step 2: Go to
Step 3: In the dialogue box that appears, select samples in columns.
Step 4: Enter ‘Jocko’ under the field marked as ‘First sample’ and ‘Other’ under the field marked as ‘Second sample’.
Step 5: In the dialogue box that appears, Click on ‘Option’. Enter 95.0 in the field of Confidence interval, 0.0 in ‘Test mean’ and ‘not equal’ in ‘Alternative’.
From Minitab result, the confidence interval is
(d)
To explain: The repayment that is recommended to the insurance company seeks.
(d)
Answer to Problem 38E
Solution: The seeking repayment is recommended between $32.2 and $195.8.
Explanation of Solution
Want to see more full solutions like this?
Chapter 7 Solutions
Introduction to the Practice of Statistics: w/CrunchIt/EESEE Access Card
- solve the question based on hw 1, 1.41arrow_forwardT1.4: Let ẞ(G) be the minimum size of a vertex cover, a(G) be the maximum size of an independent set and m(G) = |E(G)|. (i) Prove that if G is triangle free (no induced K3) then m(G) ≤ a(G)B(G). Hints - The neighborhood of a vertex in a triangle free graph must be independent; all edges have at least one end in a vertex cover. (ii) Show that all graphs of order n ≥ 3 and size m> [n2/4] contain a triangle. Hints - you may need to use either elementary calculus or the arithmetic-geometric mean inequality.arrow_forwardWe consider the one-period model studied in class as an example. Namely, we assumethat the current stock price is S0 = 10. At time T, the stock has either moved up toSt = 12 (with probability p = 0.6) or down towards St = 8 (with probability 1−p = 0.4).We consider a call option on this stock with maturity T and strike price K = 10. Theinterest rate on the money market is zero.As in class, we assume that you, as a customer, are willing to buy the call option on100 shares of stock for $120. The investor, who sold you the option, can adopt one of thefollowing strategies: Strategy 1: (seen in class) Buy 50 shares of stock and borrow $380. Strategy 2: Buy 55 shares of stock and borrow $430. Strategy 3: Buy 60 shares of stock and borrow $480. Strategy 4: Buy 40 shares of stock and borrow $280.(a) For each of strategies 2-4, describe the value of the investor’s portfolio at time 0,and at time T for each possible movement of the stock.(b) For each of strategies 2-4, does the investor have…arrow_forward
- Negate the following compound statement using De Morgans's laws.arrow_forwardNegate the following compound statement using De Morgans's laws.arrow_forwardQuestion 6: Negate the following compound statements, using De Morgan's laws. A) If Alberta was under water entirely then there should be no fossil of mammals.arrow_forward
- Negate the following compound statement using De Morgans's laws.arrow_forwardCharacterize (with proof) all connected graphs that contain no even cycles in terms oftheir blocks.arrow_forwardLet G be a connected graph that does not have P4 or C3 as an induced subgraph (i.e.,G is P4, C3 free). Prove that G is a complete bipartite grapharrow_forward
- MATLAB: An Introduction with ApplicationsStatisticsISBN:9781119256830Author:Amos GilatPublisher:John Wiley & Sons IncProbability and Statistics for Engineering and th...StatisticsISBN:9781305251809Author:Jay L. DevorePublisher:Cengage LearningStatistics for The Behavioral Sciences (MindTap C...StatisticsISBN:9781305504912Author:Frederick J Gravetter, Larry B. WallnauPublisher:Cengage Learning
- Elementary Statistics: Picturing the World (7th E...StatisticsISBN:9780134683416Author:Ron Larson, Betsy FarberPublisher:PEARSONThe Basic Practice of StatisticsStatisticsISBN:9781319042578Author:David S. Moore, William I. Notz, Michael A. FlignerPublisher:W. H. FreemanIntroduction to the Practice of StatisticsStatisticsISBN:9781319013387Author:David S. Moore, George P. McCabe, Bruce A. CraigPublisher:W. H. Freeman