Example 6.3.1: A company pays an average of $9 per hour. Asample of 21 workers shows mean wage of $8.5 and samplevariance (S2) as $0.60 Test the claim that population variance isgreater than 0.55 assuming that the sample is taken from a normalpopulation use aSolution:Step 1: Hypothesis5%.. [Using p-value method]

Answered: Image /qna-images/answer/63563760-34d4-42a9-9a12-ca072709895b.jpg

Answered: Example 6.3.1: A company pays an…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Experiment 3: Fifteen female CSCC students and 10 male CSCC students are randomly selected to determine if the mean time studying for females is significantly more than the mean study time for males. It can be assumed that all study times are normally distributed. Are these samples independent or dependent? Find the null and alternative hypothesis. H0: Ha:
Chapter Exerclse 6-84 (Algo) The Rejuvo Corp. manufactures granite countertop cleaner and polish. Quarterly sales Qis a random variable with a mean of 24,320 bottles and a standard deviation of 3,130 bottles. Variable cost is $13 per unit, and fixed cost is $220,270. (a) Find the mean and standard deviation of Rejuvo's total cost. Mean Standard deviation (b-1) If all bottles are sold, what would the selling price have to be to break even, on average? (Round your answer to 2 declmal places.) Selling price (b-2) If all bottles are sold, what would the selling price have to be to make a profit of $25,695? (Round your answer to 2 declmel places.) Selling price < Prev 2
Which statement is not correct?A. Spurious correlation can often be reduced by expressing X and Y in per capita terms.B. Autocorrelation is mainly a concern if we are using time - series data.C. Heteroscedastic residuals will have roughly the same variance for any value of X.D. Standardized residuals make it easy to identify outliers or instances of poor fit. ( don't hand writing solution)
In a production process, the time (in minutes) taken (run time) for a production run and the number of items produced (run size) for 15 randomly selected orders are analyzed using Minitab statistical software. The Minitab output is as follows: Regression Analysis: Run time versus Run size Analysis of Variance Source DF Adj SS Adj MS F-Value Critical value Regression 1 8737.1 8737.1 29.35 ……… Error 13 3870.5 297.7 Total 14 12607.6 Model Summary S R-sq R-sq(adj) R-sq(pred) 17.2549 ……. 66.94% 61.32% Coefficients Term Coef SE Coef T-Value P-Value VIF Constant 148.4 11.3 13.13 0.000 Run size 0.2627 0.0485 ……. ……. 1.00 1- Write down the least square regression equation to predict the run time for run size. 2- Interpret the coefficients of the fitted model. 3- At 5% significance level, test if run size is a good…
Exercise 4: Step 1: Test Set-up The next test is a t-test for unequal variance. Here is the problem: The human resources department at Sue, Grabitt, and Runne also tracks the cost of one-bedroom apartments in two popular neighborhoods, NoBo and SoBo. The general perception of long-time residents is that rents are probably lower in SoBo. They hope to determine whether the average rent for a one-bedroom apartment is lower in SoBo than in NoBo. The results of their survey are shown in Tables 18 and 19: Step 2. Select the Level of Significance, α A 5 percent significance has been selected. Step 3. State the Null Hypothesis (H0) and Alternate Hypothesis (H1) H0: H1: Step 4. Compose the Decision Rule Step 5. Calculate the Value of the Test Statistic, p-value, and estimate statistical power Use G*Power to calculate statistical power. Step 6. Decide and Report
An online math tutor service would like to claim that the average time spent taking a standardized test after students go through tutoring is less than the average time spent before going through tutoring. A simple random sample of students takes a pre-test, then goes through tutoring and takes a post-test. Results of the sample are shown below. Mean Variance Observations Pearson Correlation Hypothesized Mean Difference df t Stat P(T<=t) one-tail t Critical one-tail P(T<=t) two-tail t Critical two-tail Confidence Level Before After 82.333 74.944 110.118 199.820 18 0.727 18 0 17 -3.227 0.002 -1.740 0.005 -2.110 99% What are the population parameters? Pick Is the two-sample hypothesis test a paired t-test or an unpaired t-test? Pick What is the level of significance? Ex: 0.12 What is the null hypothesis? Ho: Mafter Pick Mbefore What is the alternative hypothesis? Ha : Pafter Pick Should Ho be rejected or does Ho fail to be rejected? Pick What conclusion can be drawn from the data? Pick…
Choose the appropriate statistical test. When computing, be sure to round each answer as indicated. A dentist wonders if depression affects ratings of tooth pain. In the general population, using a scale of 1-10 with higher values indicating more pain, the average pain rating for patients with toothaches is 6.8. A sample of 30 patients that show high levels of depression have an average pain rating of 7.1 (variance 0.8). What should the dentist determine? 1. Calculate the estimated standard error. (round to 3 decimals). [st.error] 2. What is thet-obtained? (round to 3 decimals). 3. What is the t-cv? (exact value) 4. What is your conclusion? Only type "Reject" or Retain"
3.Which model do you think is the “best” reduced model? Discuss why you choose this model. Analysis of Variance Table (Step back model) Response: rent Df Sum Sq Mean Sq F value Pr(>F) age 1 21000 21000 17.1136 0.0003079 *** sqft 1 35364 35364 28.8196 1.134e-05 *** sd 1 5961 5961 4.8576 0.0362339 * unts 1 8678 8678 7.0722 0.0130049 * gar 1 33364 33364 27.1899 1.713e-05 *** cp 1 7641 7641 6.2269 0.0189934 * Residuals 27 33131 1227 --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Step forward model: Response: rent Df Sum Sq Mean Sq F value Pr(>F) age 1 21000 21000 16.6255 0.000406 *** sqft 1 35364 35364 27.9976 1.757e-05 *** sd 1 5961 5961 4.7191 0.039517 * unts 1 8678 8678 6.8705 0.014697 * gar 1 33364 33364 26.4144 2.599e-05 *** cp 1 7641 7641 6.0493 0.021176 *…
Using the sample data from the accompanying table, complete parts (a) and (b). E Click the icon to view the data table (a) Explain why it does not make sense to construct confidence or prediction intervals based on the least-squares regression equation. Choose the correct answer below. O A. It does not make sense to construct confidence or prediction intervals based on the least-squares regression equation because there is a linear relationship between sugar content and calories in high-protein and moderate protein energy bars. O B. It does not make sense to construct confidence or prediction intervals based on the least-squares regression equation because there is no linear relationship between sugar content and calories in high-protein and moderate protein energy bars. O C. It does not make sense to construct confidence or prediction intervals based on the least-squares regression equation because the residuals are not normally distributed. (b) Construct a 95% confidence interval for…
The accounting department analyzes the variance of the weekly unit costs reported by two production departments. A sample of 16 cost reports for each of the two departments shows cost variances of 2.3 and 5.4, respectively. Is this sample sufficient to conclude that the two production departments differ in terms of unit cost variance? Use a = .10. a. State the null and alternative hypotheses. Ho: 0 12 equal to H : 0 1 ? not equal to 2 b. Calculate the value of the test statistic (to 2 decimals). c. The p-value is between .10 and .20 v d. What is your conclusion? Cannot conclude that the departments differ in terms of unit cost variance v
A magazine claims that the mean amount spent by a customer at Burger Stop is greater than the mean amount spent by a customer at Fry World. The results for samples of customer transactions for the two fast food restaurants are shown below. At x = 0.01, can you support the magazine's claim? Assume the population variances are equal. Assume the samples are random and independent, and the populations are normally distributed. Complete parts (a) through (e) below. Burger Stop Fry World X₁ = $11.22 X₂ = $10.59 $₁ = $0.79 $2 = $0.64 n₂ = 10 n₁ = 12 (a) Identify the claim and state Ho and H₂. Which is the correct claim below? O A. "The mean amount spent by a customer at Burger Stop is greater than the mean amount spent by a customer at Fry World." OB. "The mean amount spent by a customer at Burger Stop is equal to the mean amount spent by a customer at Fry World." O c. "The mean amount spent by a customer at Burger Stop is not equal to the mean amount spent by a customer at Fry World." O D.…

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