
To find:
To perfom the test statistic for the following scenerio,
“A manufacturer of children’s vitamins claims that its vitamins are mixed so the each batch has exactly the following percentage of each color: 20% green, 40% yellow, 10% red, and 30% orange. To test the claim that these percentage are incorrect, 100 bottles of vitamins were pulled and the clors of the vitamins were tailled”.

Answer to Problem 26E
Solution:
(a)
(b) Use the chi-square test for this goodness of fit and the level of siginificance is
(c) The test statistic value is
(d) There is sufficient evidence to support that percentages stated by the vitamin manufacturer are incorrect.
Explanation of Solution
The following table represents the random sample of number of vitamins and the color
of vitamins is,
Children's Vitamins | ||||
Green | Yellow | Red | orange | |
Number | 1149 | 1948 | 552 | 1401 |
Step 1: To state the null and alternative hypotheses.
When starting the hypotheses to be used we take the null hypothesis to be that the proportion percentages stated by the vitamin manufacturer are not incorrect.
Null hypothesis:
Alternative hypothesis:
Let
Mathematically, we can write the null and alternative hypothesis for four different vitamins as follows,
Step 2: To determine whether the observed values proportion of color of vitamins is match with the
Step 3: To calculate the test statistic.
Let‘s calculate the expected value of color of vitamins. Since we are assuming that the proportion percentages stated by the vitamin manufacturer are not incorrect is calculated as follows,
And,
Formula for calculating test statistic for a Chi-Square test for goodness of fit:
The test statistic for a chi-square test for goodness of fit is given below,
Where Oi, is the observed frequency for the ith possible outcome and Ei is the expected frequency for the ith possible outcome and n is the sample size.
From the above table represent the value of n is 5050.
The following table represents the test statistic,
Children's Vitamins | Observed Values | Expected Values | |||
Green | 1149 | 1010 | 139 | 19321 | 19.1297 |
Yellow | 1948 | 2020 | -72 | 5184 | 2.56634 |
Red | 552 | 505 | 47 | 2209 | 4.37426 |
Orange | 1401 | 1515 | -114 | 12996 | 8.57822 |
Step 4: Degrees of freedom in a Chi-square test for goodness of fit:
In a chi-square test for goodness of fit the number of degrees of freedom for the chi-square distribution of the test statistic is given by,
Where k is the number of possible outcomes for each trial.
Rejection Region for Chi-Square Test for Association:
Reject the null hypothesis,
From the given information k = 4.
Substitute the above values in the formula of degrees of freedom to get the following,
Conclusion:
Use the level of significance of
Use the “Area to the right of the critical value
The test statistic value is
By using the above condition we reject the null hypothesis.
Since,
So, there is sufficient evidence to support that percentages stated by the vitamin manufacturer are incorrect.
Final statement:
(a)
(b) Use the chi-square test for this goodness of fit and the level of siginificance is
(c) The test statistic value is
(d) There is sufficient evidence to support that percentages stated by the vitamin manufacturer are incorrect.
Want to see more full solutions like this?
Chapter 10 Solutions
Beginning Statistics, 2nd Edition
- 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
- 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





