Treatment Treatment TreatmentABсс324532304435304534264735324939Sample mean304635Sample variance6.004.006.50a. At the α = 0.05 level of significance, can we reject the null hypothesis that the means of the three treatments are equal?Compute the values below (to 1 decimal, if necessary).Sum of Squares, TreatmentSum of Squares, ErrorMean Squares, TreatmentMean Squares, ErrorCalculate the value of the test statistic (to 2 decimals).The p-value is - Select your answer -What is your conclusion?- Select your answer -b. Calculate the value of Fisher's LSD (to 2 decimals).Use Fisher's LSD procedure to test whether there is a significant difference between the means for treatments A and B, treatments A and C, and treatments B and C. Use a = 0.05.Difference Absolute ValueConclusion- Select your answer -TA-TBTA-ECTB-TCc. Use Fisher's LSD procedure to develop a 95% confidence interval estimate of the difference between the means of treatments A and B (to 2 decimals). Enter negative values as negative numbers.- Select your answer -Select your answer -

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MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

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H1 H2 A study was done using a treatment group and a placebo group. The results are shown in the table. Assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. Complete parts (a) and (b) below. Use a 0.10 significance level for both parts. 25 36 2.37 2.69 s 0.69 0.96 a. Test the claim that the two samples are from populations with the same mean. What are the null and alternative hypotheses? O A. Ho: H1 <#2 H: H1 2 42 VB. Họ: H1 = P2 H: 1 42 O C. Ho: H1 #H2 O D. Ho: H1 = H2 H: 41
Evaluation 1. Samples of size n were randomly selected from the population with means and variances given below. In each case, find the mean, variance, and the variance of the sampling distribution of the means. a. n = 49; μ = 25; o² = 6.25 b. n = 55; μ = 10; o² = 9 2. A random sample of n = 100 measurements is obtained from a population with μ = 55 and o = 20. Describe the sampling distribution of means by computing the (a) mean of the sampling distribution of means μ, (b) variance of the sampling distribution of means o, and the (c) standard deviation of the sampling distribution of means o.
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An experiment with a single factor has been conducted as a completely randomized design. A portion of the output is shown below. Source MS F DF 3 25.69 Factor 84.35 Error ? Total 161.42 15 The estimate of the variance of any observation in the experiment is 7.03
Federal Government Employee E-mail Use It has been reported that 73% of federal government employees use e-mail. If a sample of 190 federal government employees is selected, find the mean, variance, and standard deviation of the number who use e-mail. Round your answers to three decimal places. Part: 0/ 2 Part 1 of 2 (a) Find the mean. Mean: μ %3D Part: 1/2 Part 2 of 2 (b) Find the variance and standard deviation. Variance: o = Standard deviation: o =
Teacher Tutorial Rate(in thousand)x A 8 B 12 16 20 C D E 24 F 28 SHOW YOUR COMPUTATION: (handwritten) 1. Solve for the mean of the population u. 2. Solve for the mean of the sampling distribution of the sample means lg- 3. Compare u and ug. 4. Solve for the variance (o2) and the standard deviation (6) of the population. 5. Solve the variance (os) and the standard deviation (oz) of the sampling distribution of the sample means µg. 6. Compare o and og.
Because variance divided by the sample size gives you the estimated standard error, as the sample size ______, the estimated standard error will ______. a. increases ; decrease b. decreases ; decrease c. increases ; increase d. decreases ; increase
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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 *…
STATISTICAL ANALYSIS (No long explanation needed. Rate will be given and write the complete solutions.) CI on Mean, Variance Unknown, Prediction and Tolerance Interval A CNC (computer numerical control) machine produces iron automobile crankshafts. Samples are measured and the inner diameter are found to be 1.01, 0.97, 1.03, 1.04, 0.99, 0.98, 0.99, 1.01, and 1.03 centimeters. For all computations, assume an approximately normal distribution. The sample mean and standard deviation for the given data are = 1.0056, and s = 0.0246. Find a 99% confidence interval on the mean of diameter. Compute a 99% prediction interval on a measured diameter of a single crankshaft piece taken from the machine. Find the 99% tolerance limits that will contain most of the metal pieces produced by the CNC machine. Choose the correct answer. a. CI on the mean: 0.970 < μ < 1.9845. Prediction Interval: 0.9186 ≤ Xn+1 ≤ 1.0926. Tolerance Interval (0.8937 and 1.1175) b. CI on the mean: 0.9781 < μ <…

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Sample variance 6.00 a. At the a= 0.05 level of significance, can we reject the null hypothesis that the means of the three treatments are equal? Compute the values below (to 1 decimal, if necessary). Sum of Squares, Treatment Sum of Squares, Error Mean Squares, Treatment. 4.00 6.50