6. Two methods for teaching reading were applied to two randomly selected groups of elementary schoolchildrenand were compared on the basis of a reading comprehension test given at the end of the learning period. Thesample means and variances computed from the test scores and shown in the accompanying table.StatisticMethod 1Method 211146469s25271(a) Find a 98% confidence interval for u1 – µ2 assuming normal populations with unknown but equalvariances o? and o?.(b) Find a 98% confidence interval for u1 - µ2 assuming normal populations with unknown variances o?and o? but known o3/o? = 1.5. Hint: use the formula of the above Problem 5.(c) (Bonus) Find a 95% confidence interval for o?/o. What do you conclude based on this result?

Given information Confidence interval is given by

Answered: 6. Two methods for teaching reading…

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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Using traditional methods it takes 8.8 hours to receive a basic driving license. A new license training method using Computer-Aided Instruction (CAI) has been proposed. A researcher used the technique on 17 students and observed that they had a mean of 8.4 hours with a variance of 2.89. Is there evidence at the 0.1 level that the technique performs differently than the traditional method? Assume the population distribution is approximately normal. Step 4 of 5 :   Determine the decision rule for rejecting the null hypothesis. Round your answer to three decimal places.
Given two independent random samples with the following results: n1=12x‾1=73s1=27n1=12x‾1=73s1=27 n2=12x‾2=102s2=22n2=12x‾2=102s2=22 Use this data to find the 95%95% confidence interval for the true difference between the population means. Assume that the population variances are equal and that the two populations are normally distributed. Step 1 of 3 :   Find the point estimate that should be used in constructing the confidence interval.
A study was performed concerning risk factors for carotid-artery stenosis (arterial narrowing) among 264 men. The data reported is:   Sample size Mean Standard deviationGroup 1: No stenosis 156 5.3 1.4Group 2: Stenosis 108 5.1 0.8 The test stastistic F is assumed to be 3.06 to check the equal variance. Question: Suppose researchers want to test if there is a significant difference in mean blood-glucose level between men with and without stenosis considering the critical F test is 1.35 with a equal variance (pooled), what is the test statistic for this test ?
Two different types of catalysts are used in a chemical process. We want to conduct a hypothesis test to see if there is a difference in the yields at 5% significance level. Assume both populations are normaly distributed. To this end, we collect a simple random sample from each process. Assume equal variances. = = n-11, n=13,x,-90, x 91.4, 8, 4, 8=4.5
When computing the variance using a sample of size n = 30, we divide the sum of squared deviations by Give your answer as a whole number.
Gestation period is the length of pregnancy, or to be more precise, the interval between fertilization and birth. In Syrian hamsters, the average gestation period is 16 days. Suppose you have a sample of 31 Syrian hamsters who were exposed to high levels of the hormone progesterone when they were pups, and who have an average gestation length of 17.1 days and a sample variance of 26.0 days. You want to test the hypothesis that Syrian hamsters who were exposed to high levels of the hormone progesterone when they were pups have a different gestation length than all Syrian hamsters. Calculate the t statistic. To do this, you first need to calculate the estimated standard error. The estimated standard error is sMM= a. 31, b. 0.5232, c. 0.7328, d. 0.9158. The t statistic is- a. 1.50, b. 2.10, c. 2.63, d. 1.20 Now suppose you have a larger sample size n = 95. Calculate the estimated standard error and the t statistic for this sample with the same sample average and the same…
Two groups of individuals were compared with respect to a high carbohydrate low- fat diet (LF) and a high-fat low carbohydrate diet (LC). Mood was assessed using a total mood disturbance score where a lower score is associated with a less negative mood. Assuming that the random samples come from independent normal distributions with a common but unknown variance, construct a 95% confidence interval for the difference between the two population means, LC less LF. Group LC LF 28+22.37 28+18.56 28+22.30 28±8.96 n 12 13 mean 47.3 19.3 standard deviation 28.3 25.8
Three new different models of compact SUVs have just arrived at the market. You are interested in comparing the gas mileage performance of all three models to see if they are the same. A partial computer output for twelve compact SUVs of each model is given below: You are told that the gas mileage population distributions for all three models are not normally distributed. What is your conclusion on the test using a 5% level of significance? There is not enough evidence of a difference in mean gas mileage. There is evidence of a difference in mean gas mileage. There is not enough evidence of a difference in median gas mileage. There is evidence of a difference in median gas mileage.
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A study was conducted to determine if differences in new textbook prices exist between on-campus bookstores, off-campus bookstores, and Internet bookstores. To control for differences in textbook prices that might exist across disciplines, the study randomly selected 12 textbooks and recorded the price of each of the 12 books at each of the three retailers. You may assume normality and equal-variance assumptions have been met. The partially completed ANOVA table based on the study's findings is shown here: ANOVA Source of Variation Textbooks Retailer Error Total SS 16,624 2.4 df MS F 17,477.6 Based on the study's findings, was it correct to block for differences in textbooks? Conduct the appropriate test at the alpha = 0.10 level of significance.

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