
Automatic Transmissions
Assume the following data values are from a 2016 Auto Guide. The guide compared various parameters of U.S.- and foreign-made cars. This report centers on the price of an optional automatic transmission. Which country has the greater variability in the price of automatic transmissions? Answer the following questions.
1. What is the null hypothesis?
2. What test statistic is used to test for any significant differences in the variances?
3. Is there a significant difference in the variability in the prices between the German cars and the U.S. cars?
4. What effect does a small
5. What degrees of freedom are used for the statistical test?
6. Could two sets of data have significantly different variances without having significantly different means?
1.

To write: The null hypothesis.
Answer to Problem 1AC
The null hypothesis is,
Explanation of Solution
Given info:
The data shows that the data values from a 2016 Auto Guide. The guide compared various parameters of U.S.- and foreign-made cars and greater variability in the price of automatic transmissions for the countries is tested.
Justification:
Here, it is observed that the greater variability in the price of automatic transmissions for the countries is tested.
Thus, the null hypothesis can be written as,
2.

To identify: The test is used to test for any significant difference in the variances.
Answer to Problem 1AC
The test is used to test for any significant difference in the variances is F-test.
Explanation of Solution
Justification:
F-test:
The F-test is used to test for the significant difference in the variances.
Here, it is observed that F-test is used to test for any significant difference in the variances.
3.

To test: Whether there is a significant difference in the variability in the prices between the German cars and the U.S. cars.
Answer to Problem 1AC
There is no sufficient evidence to support the claim that there is a significant difference in the variability in the prices between the German cars and the U.S. cars.
Explanation of Solution
Justification:
Step by step procedure for finding the sample variances using Minitab procedure is,
- Choose Stat > Basic Statistics > Display Descriptive Statistics.
- In Variables enter the columns Germany (2016) and U.S. Cars (2016).
- Choose option statistics, and select Variance.
- Click OK.
Output using Minitab procedure is,
Here, the sample variances
Where,
Here, the variability in the price of automatic transmissions for the countries is tested. Hence, the claim is that,
Test statistic:
The formula to find the F-statistic is,
Substitute
Thus, the F-statistic value is 1.090.
Degrees of freedom:
The degrees of freedom are
That is,
Critical value:
The level of significance is
Divide the level of significance by 2.
That is,
The critical F-value for a two-tailed test is obtained using the Table H: The F-Distribution with the level of significance 0.025.
Procedure:
- Locate 3 in the degrees of freedom, denominator row of the Table H.
- Obtain the value in the corresponding degrees of freedom, numerator column below 3.
That is,
Rejection region:
The null hypothesis would be rejected if
Conclusion:
Here, the F-value is lesser than the critical value.
That is,
Thus, it can be concluding that the null hypothesis is not rejected.
Hence, there is no sufficient evidence to support the claim that there is a significant difference in the variability in the prices between the German cars and the U.S. cars.
4.

The effect of the small sample sizes have on the standard deviations.
Answer to Problem 1AC
The effect of the small sample sizes have on the standard deviations is the small sample sizes are impacted by outliers.
Explanation of Solution
Justification:
Here, it is observed that taking the small sample sizes gives the do not give appropriate results because the small sample sizes are impacted by outliers. Thus, it can be conclude that effect of the small sample sizes have on the standard deviations is the small sample sizes are impacted by outliers.
5.

To find: The degrees of freedom are used for the statistical test.
Answer to Problem 1AC
The degrees of freedom are used for the numerator and denominator both 3.
Explanation of Solution
Justification:
Form the part (3), it can be observed that thedegrees of freedom are used for the numerator and denominator 3.
6.

To check: Whether the two sets of data have significantly different variances without having significantly different means.
Answer to Problem 1AC
Yes, the two sets of data have significantly different variances without having significantly different means.
Explanation of Solution
Justification:
From the given information, it can be observed that the mean of the two sets of data is same in the center but the standard deviations of the two sets are different. Thus, it can be conclude that the two sets of data have significantly different variances without having significantly different means.
Want to see more full solutions like this?
Chapter 9 Solutions
Elementary Statistics: A Step By Step Approach
- For a binary asymmetric channel with Py|X(0|1) = 0.1 and Py|X(1|0) = 0.2; PX(0) = 0.4 isthe probability of a bit of “0” being transmitted. X is the transmitted digit, and Y is the received digit.a. Find the values of Py(0) and Py(1).b. What is the probability that only 0s will be received for a sequence of 10 digits transmitted?c. What is the probability that 8 1s and 2 0s will be received for the same sequence of 10 digits?d. What is the probability that at least 5 0s will be received for the same sequence of 10 digits?arrow_forwardV2 360 Step down + I₁ = I2 10KVA 120V 10KVA 1₂ = 360-120 or 2nd Ratio's V₂ m 120 Ratio= 360 √2 H I2 I, + I2 120arrow_forwardQ2. [20 points] An amplitude X of a Gaussian signal x(t) has a mean value of 2 and an RMS value of √(10), i.e. square root of 10. Determine the PDF of x(t).arrow_forward
- In a network with 12 links, one of the links has failed. The failed link is randomlylocated. An electrical engineer tests the links one by one until the failed link is found.a. What is the probability that the engineer will find the failed link in the first test?b. What is the probability that the engineer will find the failed link in five tests?Note: You should assume that for Part b, the five tests are done consecutively.arrow_forwardProblem 3. Pricing a multi-stock option the Margrabe formula The purpose of this problem is to price a swap option in a 2-stock model, similarly as what we did in the example in the lectures. We consider a two-dimensional Brownian motion given by W₁ = (W(¹), W(2)) on a probability space (Q, F,P). Two stock prices are modeled by the following equations: dX = dY₁ = X₁ (rdt+ rdt+0₁dW!) (²)), Y₁ (rdt+dW+0zdW!"), with Xo xo and Yo =yo. This corresponds to the multi-stock model studied in class, but with notation (X+, Y₁) instead of (S(1), S(2)). Given the model above, the measure P is already the risk-neutral measure (Both stocks have rate of return r). We write σ = 0₁+0%. We consider a swap option, which gives you the right, at time T, to exchange one share of X for one share of Y. That is, the option has payoff F=(Yr-XT). (a) We first assume that r = 0 (for questions (a)-(f)). Write an explicit expression for the process Xt. Reminder before proceeding to question (b): Girsanov's theorem…arrow_forwardProblem 1. Multi-stock model We consider a 2-stock model similar to the one studied in class. Namely, we consider = S(1) S(2) = S(¹) exp (σ1B(1) + (M1 - 0/1 ) S(²) exp (02B(2) + (H₂- M2 where (B(¹) ) +20 and (B(2) ) +≥o are two Brownian motions, with t≥0 Cov (B(¹), B(2)) = p min{t, s}. " The purpose of this problem is to prove that there indeed exists a 2-dimensional Brownian motion (W+)+20 (W(1), W(2))+20 such that = S(1) S(2) = = S(¹) exp (011W(¹) + (μ₁ - 01/1) t) 롱) S(²) exp (021W (1) + 022W(2) + (112 - 03/01/12) t). where σ11, 21, 22 are constants to be determined (as functions of σ1, σ2, p). Hint: The constants will follow the formulas developed in the lectures. (a) To show existence of (Ŵ+), first write the expression for both W. (¹) and W (2) functions of (B(1), B(²)). as (b) Using the formulas obtained in (a), show that the process (WA) is actually a 2- dimensional standard Brownian motion (i.e. show that each component is normal, with mean 0, variance t, and that their…arrow_forward
- The scores of 8 students on the midterm exam and final exam were as follows. Student Midterm Final Anderson 98 89 Bailey 88 74 Cruz 87 97 DeSana 85 79 Erickson 85 94 Francis 83 71 Gray 74 98 Harris 70 91 Find the value of the (Spearman's) rank correlation coefficient test statistic that would be used to test the claim of no correlation between midterm score and final exam score. Round your answer to 3 places after the decimal point, if necessary. Test statistic: rs =arrow_forwardBusiness discussarrow_forwardBusiness discussarrow_forward
- Glencoe Algebra 1, Student Edition, 9780079039897...AlgebraISBN:9780079039897Author:CarterPublisher:McGraw HillBig Ideas Math A Bridge To Success Algebra 1: Stu...AlgebraISBN:9781680331141Author:HOUGHTON MIFFLIN HARCOURTPublisher:Houghton Mifflin HarcourtHolt Mcdougal Larson Pre-algebra: Student Edition...AlgebraISBN:9780547587776Author:HOLT MCDOUGALPublisher:HOLT MCDOUGAL


