6. The three samples below have been obtained from normal populations withequal variances. Test the hypothesis that the sample means are equal:81210710131412119.14The table value of Fat 5% level of significance for v = 2 and v2 = 12 is 3-88.

To compare the population means of more than 2 groups the ANOVA F single factor test is used. This…

Answered: 6. The three samples below have been…

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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The following information is available for two samples selected from independent normally distributed populations. Population A: n1=30 S2/1=25 Population B: n2=30 S2/2=36 a. Which sample variance do you place in the numerator of FSTAT? b. What is the value of FSTAT?
You are testing the claim that having lights on at night increases weight gain (abstract). A sample of 10 mice lived in an environment with bright light on all of the time and 8 mice who lived in an environment with a normal light/dark cycle is given below. Test the claim using a 10% level of significance. Assume the population variances are unequal and that the weight changes are normally distributed. Give answers to 3 decimal places. Data available at StatKey, choose Mice Wgt Gain-2e data set Light (x₁) 1.71 4.67 4.99 5.33 5.43 6.94 7.15 9.17 10.26 11.67 What are the correct hypotheses? Note this may view better in full screen mode. Select the correct symbols in the order they appear in the problem. Ho: Select an answer ? V Select an answer V Ha: Select an answer ✓ Based on the hypotheses, find the following: Test Statistic = Dark (₂) 2.27 2.53 2.83 4 4.21 4.6 5.95 6.52 p-value = ? V Select an answer V (Hint: difference in means from Ha) The correct decision is to Select an answer…
. 3. A researcher is interested in knowing if preterm infants with late metabolic acidosis and preterm infants without the condition differ with respect to urine levels of a certain chemical. The mean levels, standard deviations, and sample sizes from the two samples are given below. Assuming that the populations have equal variances: a. Fill in the missing pieces of the Stata output. ttesti 38 8.5 3.5 32 6.85 2.6 X y Two-sample t test with A variances combined diff Obs Ho: diff=0 38 32 Sample With condition 70 diff mean(x) n 38 Without condition 32 Ha: diff |t|) = F [95% Conf. Intervall 7.349579 5.9126 6.98079 .1552544 t = degrees of freedom = 9.650421 7.7874 8.510638 3.144746 E b. What is the conclusion of your test? c. Use this output to construct a 99% confidence interval for the difference between the two means. 68 Ha: diff0 Pr(Tt) = 0.0155
A medical researcher wants to test whether the mean cholesterol levels are the same for 3 treatment groups of subjects (with 18 total randomly selected subjects): those who take a new medication, those who follow a diet, and those who make no changes to medication or diet. They plan to use a 5% level of significance for their decision. Below is a partial Analysis of Variance table. Use it to answer the following questions deg of freedom Sum of Squares Mean Square F statistic Treatment Group 3.121 1.5607 Error 1.517 0.1011 Total 17 What are the degrees of freedom for the treatment groups? What are the degrees of freedom for the error? What is the F-statistic for this data (3 decimal places)
For a repeated-measures study comparing two treatment conditions, a researcher obtains a sample of n = 10 difference scores with a mean of MD = 6 and a variance of s2 = 90. What is the value for the repeated-measures t statistic for these data? Select one: a. 7 b. 6 c. 2 d. 15
4 4G I. moodle.unizwa.edu.om a | Quiz3 Time left 0:00:21 Question 3 Answer saved Marked out of 2.00 P Flag question Which one of the following choices describes a problem for which an analysis of variance would be appropriate? O A. Comparing the proportion of successes for three different treatments of anxiety. Each treatment is tried on 100 patients O B. Analyzing the relationship between high school GPA and college GPA O . Analyzing the relationship between gender and opinion about capital punishment (favor or oppose). O D. Comparing the mean birth weights of newborn babies for three different hospitals. O E. None of these Clear my choice Previous page Next page 1 2w-ANOVA(3) Jump to... Quiz navigation 7 +
3.18. A manufacturer of television sets is interested in the effect on tube conductivity of four different types of coating for color picture tubes. A completely randomized experiment is conducted and the following conductivity data are obtained: Coating Type 1 234 2 4 143 152 134 129 Conductivity 150 137 132 132 141 149 136 127 146 143 127 129 (a) Is there a difference in conductivity due to coating type? Use a = 0.05. (b) Estimate the overall mean and the treatment effects. (c) Compute a 95 percent confidence interval estimate of the mean of coating type 4. Compute a 99 percent confidence interval estimate of the mean difference between coating types 1 and 4.
1. The tables below show the results of an independent samples t-test in SPSS conducted using the SPSS data we have been using this semester. Two groups ("Yes" and "No") are compared on their "final". Yes and no refer to whether they attended the review sessions or not. Review the results carefully and answer the questions below. (12 points) final final Attended review sessions? No Yes Equal variances assumed Equal variances not assumed Levene's Test for Equality of Variances F Group Statistics 219 Sig. N .641 35 70 t -2.693 d. What are the df for the t-test we use? 103 Mean Independent Samples Test -2.719 59.83 64.14 Std. Deviation 7.591 7.810 df Sig. (2-tailed) 103 .008 69.872 .008 e. Are the groups significantly different from each other? f. Which group scored higher? Yes Group No Group The two groups are not significantly different_ t-test for Equality of Means Mean Difference -4.314 Std. Error Mean -4.314 1.283 .933 Std. Error Difference 1.602 1.587 95% Confidence Interval of the…
Which of the following factors have each of the advantages listed when using analysis of variance? Multiple answers can be chosen. 1. Makes ANOVA more robust to departures from the assumption of normality. 2. Makes ANOVA more robust to departures from the assumption of equal variance. 3. Increases the power of the test. Options: Large sample size. Balanced design.
What is the value of n (the number of observations)? What is the value of Se (standard error for the model) (2 decimals)? What is the average of the x variable (2 decimals)? What is the variance of the x variable (1 decimal)?
5. We conduct a two-sample t-test to compare the means of two populations. Assume that the populations have equal variance. Two random samples are taken from the populations and the following summary statistics was obtained. Descriptive Statistics: Y1, Y2 Total Variable Count Mean SE Mean StDev Variance Sum Yl 16 20.19 1.31 5.24 27.46 323.04 16 24.42 1.97 7.87 61.94 414.72 Y2 Test the hypothesis Ho: H1 = Hz versus the one-sided alternative H: H < Hz at a = 0.01 level. State %3D a. your conclusion. Determine the bounds on the p-value of the test. Construct a 95% upper confidence bound on the difference in means. b. c.
1.com Statistics students believe that the mean score on a first statistics test is 65. The instructor thinks that the mean score is higher. She samples 10 statistics students and obtains the scores: Grades 96 69 83.2 69 65 85.5 74.4 65 85.5 66.5 Test grades are believed to be normally distributed. Use a significance level of 5%. A. State the alternative hypothesis: H: Ομ 65 B. State the mean of the sample: 75.91 (Round to two decimal places.) C. State the standard error the sample means: 3.440 (Round to four decimal places.) D. State the test statistic: t = (Round to four decimal places.) E. State the p-value: (Round to four decimal places.) F. Decision: O Do not reject the null hypothesis. O Reject the null hypothesis.

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6. The three samples below have been obtained from normal populations with equal variances. Test the hypothesis that the sample means are equal: 8. 7 12 10 7 10 13 14 12 11 9. 14 The table value of Fat 5% level of significance for vi = 2 and v2 = 12 is 3-88.