Concept explainers
a.
To identify: The claim and state
a.
![Check Mark](/static/check-mark.png)
Answer to Problem 16CQ
The claim is that “there is a significant difference in price”.
The hypotheses are given below:
Null hypothesis:
Alternative hypothesis:
Explanation of Solution
Given info:
Justification:
Here, there is a significant difference in prices tested. Hence, the claim is that there is a significant difference in price. This can be written as
The hypotheses are given below:
Null hypothesis:
Alternative hypothesis:
b.
To find: The critical value
b.
![Check Mark](/static/check-mark.png)
Answer to Problem 16CQ
The critical value at
Explanation of Solution
Calculation:
Here, the test is two tailed test.
Critical value:
Here, variances are not equal. Hence, the degrees of freedom is,
Software Procedure:
Step-by-step procedure to obtain the critical value using the MINITAB software:
- Choose Graph >
Probability Distribution Plot choose View Probability> OK. - From Distribution, choose ‘t’ distribution.
- In Degrees of freedom, enter 11.
- Click the Shaded Area tab.
- Choose Probability value and Two Tail for the region of the curve to shade.
- Enter the Probability value as 0.01.
- Click OK.
Output using the MINITAB software is given below:
From the output, the critical value is
c.
To find: The test value.
c.
![Check Mark](/static/check-mark.png)
Answer to Problem 16CQ
The test value is 11.09.
Explanation of Solution
Calculation:
Test statistic:
Software Procedure:
Step by step procedure to obtain test statistic using the MINITAB software:
- Choose Stat > Basic Statistics > 2-Sample t.
- Choose Summarized data.
- In first, enter
Sample size as 12, Mean as 1.43, Standard deviation as 0.09. - In second, enter Sample size as 16, Mean as 1.03, Standard deviation as 0.1.
- Choose Options.
- In Confidence level, enter 99.
- In Alternative, select not equal.
- Click OK in all the dialogue boxes.
Output using the MINITAB software is given below:
From the MINITAB output, the test value is 11.09.
d.
To make: The decision.
d.
![Check Mark](/static/check-mark.png)
Answer to Problem 16CQ
The decision is that, the null hypothesis isrejected.
Explanation of Solution
Calculation:
Software Procedure:
Step-by-step procedure to indicate the appropriate area and critical value using the MINITAB software:
- Choose Graph > Probability Distribution Plot choose View Probability> OK.
- From Distribution, choose ‘t’ distribution.
- In Degrees of freedom, enter 11.
- Click the Shaded Area tab.
- Choose Probability value and Two Tail for the region of the curve to shade.
- Enter the Probability value as 0.01.
- Enter 11.09 under show reference lines at X values
- Click OK.
Output using the MINITAB software is given below:
From the output, it can be observed that the test statistic value falls in the critical region. Therefore, the null hypothesis isrejected.
e.
To summarize: The result.
To construct: The 99% confidence interval for the difference of the mean.
e.
![Check Mark](/static/check-mark.png)
Answer to Problem 16CQ
The conclusion is that, there is enough evidence to support the claim that there is a significant difference in price.
The 99% confidence interval for the difference of the mean is
Explanation of Solution
Justification:
From part (d), the null hypothesis is rejected. Thus, there is enough evidence to support the claim that there is a significant difference in price.
Confidence interval:
From the MINITAB output in part (c), the 99% confidence interval for the difference of the mean is
Want to see more full solutions like this?
Chapter 9 Solutions
ELEMENTARY STATISTICS: STEP BY STEP- ALE
- I need help with this problem and an explanation of the solution for the image described below. (Statistics: Engineering Probabilities)arrow_forward310015 K Question 9, 5.2.28-T Part 1 of 4 HW Score: 85.96%, 49 of 57 points Points: 1 Save of 6 Based on a poll, among adults who regret getting tattoos, 28% say that they were too young when they got their tattoos. Assume that six adults who regret getting tattoos are randomly selected, and find the indicated probability. Complete parts (a) through (d) below. a. Find the probability that none of the selected adults say that they were too young to get tattoos. 0.0520 (Round to four decimal places as needed.) Clear all Final check Feb 7 12:47 US Oarrow_forwardhow could the bar graph have been organized differently to make it easier to compare opinion changes within political partiesarrow_forward
- 30. An individual who has automobile insurance from a certain company is randomly selected. Let Y be the num- ber of moving violations for which the individual was cited during the last 3 years. The pmf of Y isy | 1 2 4 8 16p(y) | .05 .10 .35 .40 .10 a.Compute E(Y).b. Suppose an individual with Y violations incurs a surcharge of $100Y^2. Calculate the expected amount of the surcharge.arrow_forward24. An insurance company offers its policyholders a num- ber of different premium payment options. For a ran- domly selected policyholder, let X = the number of months between successive payments. The cdf of X is as follows: F(x)=0.00 : x < 10.30 : 1≤x<30.40 : 3≤ x < 40.45 : 4≤ x <60.60 : 6≤ x < 121.00 : 12≤ x a. What is the pmf of X?b. Using just the cdf, compute P(3≤ X ≤6) and P(4≤ X).arrow_forward59. At a certain gas station, 40% of the customers use regular gas (A1), 35% use plus gas (A2), and 25% use premium (A3). Of those customers using regular gas, only 30% fill their tanks (event B). Of those customers using plus, 60% fill their tanks, whereas of those using premium, 50% fill their tanks.a. What is the probability that the next customer will request plus gas and fill the tank (A2 B)?b. What is the probability that the next customer fills the tank?c. If the next customer fills the tank, what is the probability that regular gas is requested? Plus? Premium?arrow_forward
- 38. Possible values of X, the number of components in a system submitted for repair that must be replaced, are 1, 2, 3, and 4 with corresponding probabilities .15, .35, .35, and .15, respectively. a. Calculate E(X) and then E(5 - X).b. Would the repair facility be better off charging a flat fee of $75 or else the amount $[150/(5 - X)]? [Note: It is not generally true that E(c/Y) = c/E(Y).]arrow_forward74. The proportions of blood phenotypes in the U.S. popula- tion are as follows:A B AB O .40 .11 .04 .45 Assuming that the phenotypes of two randomly selected individuals are independent of one another, what is the probability that both phenotypes are O? What is the probability that the phenotypes of two randomly selected individuals match?arrow_forward53. A certain shop repairs both audio and video compo- nents. Let A denote the event that the next component brought in for repair is an audio component, and let B be the event that the next component is a compact disc player (so the event B is contained in A). Suppose that P(A) = .6 and P(B) = .05. What is P(BA)?arrow_forward
- Glencoe Algebra 1, Student Edition, 9780079039897...AlgebraISBN:9780079039897Author:CarterPublisher:McGraw Hill
![Text book image](https://www.bartleby.com/isbn_cover_images/9780079039897/9780079039897_smallCoverImage.jpg)