The following information was obtained from an experiment with two treatments (A and B).DATA:nMeanVarianceTreatment A104.2332.14Treatment B53.243.05The researcher wants to evaluate the homoscedasticity of the data by evaluating the variance ratio (0²₁/ 0²2), andcalculating the lower and upper confidence interval of the variance ratio.Using a 98% Confidence level:1. What is the upper critical value of F (F2 or Fa/2). You can use the F-table provided in class or Excel (round to the nearesthundredth)2. What is the upper confidence level of the variance ratio (round to the nearest hundredth)3. Evaluate the confidence interval and indicate if the variances are (yes) or are not (no) homogeneous (yes/no, choose one)

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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

Problem 1P

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The table below lists weights (carats) and prices (dollars) of randomly selected diamonds. Find the (a) explained variation, (b) unexplained variation, and (c) indicated prediction interval. There is sufficient evidence o support a claim of a linear correlation, so it is reasonable to use the regression equation when making predictions. For the prediction interval, use a 95% confidence level with a diamond that weighs 0.8 carats. Weight Price 0.3 $508 a. Find the explained variation. 0.4 $1153 0.5 $1332 Round to the nearest whole number as needed.) . Find the unexplained variation. Round to the nearest whole number as needed.) c. Find the indicated prediction interval.
A paint manufacturer wants to introduce a new paint on the market. The following data were obtained for drying time (minutes) when five samples of the paint were tested. 380 290 240 310 300 Assuming that the drying time is normally distributed with a variance of 400, obtain a 99% confidence interval for the true mean drying time.
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 researcher was interested in knowing if mean BMI was different between males and females. He obtained the following test results. Choose the correct interpretation of the results: F test p value = 0.30 Two sample t test, assume equal variance = 0.02 Two sample t test, assume unequal variance = 0.61 Mean BMI is not significantly different between males and females. None Mean BMI is higher in males than females. Mean BMI is significantly different between males and females.
2 different treatments were used to speed the relief time (in minutes) of migraines. 8 subjects received each treatment. The recovery times are listed below. Treatment 1: 20 19 29 25 34 32 39 38 Treatment 2: 18 18 16 17 29 28 21 37 Construct a 90% confidence interval for the difference in the relief time between the 2 treatments.
Using traditional methods it takes 8.2 hours to receive a basic flying license. A new license training method using Computer Aided Instruction (CAI) has been proposed. A researcher used the technique on 26 students and observed that they had a mean of 8.0 hours with a variance of 2.89. Is there evidence at the 0.1 level that the technique reduces the training time? Assume the population distribution is approximately normal. Step 3 of 5: Specify if the test is one-tailed or two-tailed.
In September a biological research team caught, weighed, and released a random sample of 24 chipmunks in Rocky Mountain National Park. The mean of the sample weights was x-bar = 8.7 oz. with sample standard deviation s = 1.4 oz. Find a 90% confidence interval for the mean September weight of all chipmunks in Rocky Mountain National Park.
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.
An intercity bus company is attempting to improve the reliability of its service. One manager claims that the variability (i.e., standard deviation) in arrivals from the posted schedule at the central bus depot is approximately 3 minutes. Another manager disagrees, suspecting that the spread in times of arrival is different. Using variance, what is the population parameter under consideration? What hypotheses should the managers test in this situation?
In a study performed on comfort food eating during times of stress, it was found that mean rated comfort food consumption level (on a scale from 1-20) for a group of students undertaking their exam period was 14. In comparison, a group of students who were not undertaking exams or examination preparation had a mean rated comfort food consumption level of 10. The pooled estimate of the population variance was 64. What is the estimated effect size (d) of the study? 0.80 0.06 0.50 2.00
Using traditional methods it takes 8.2 hours to receive a basic flying license. A new license training method using Computer Aided Instruction (CAI) has been proposed. A researcher used the technique on 26 students and observed that they had a mean of 8.0 hours with a variance of 2.89. Is there evidence at the 0.1 level that the technique reduces the training time? 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.
An automobile manufacturer has given its van a 50.6 miles/gallon (MPG) rating. An independent testing firm has been contracted to test the actual MPG for this van since it is believed that the van has an incorrect manufacturer's MPG rating. After testing 220 vans, they found a mean MPG of 50.5. Assume the population variance is known to be 1.69. A level of significance of 0.01 will be used. Find the P-value of the test statistic. you may write the P-value as a range using interval notation, or as a decimal value rounded to four decimal places.

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