Y is regressed against X1, X2, X3, X4, Xs, X6, X7, and Xg. Consider the following test:Ho: B1 = B2 = B3 = B4 = B5 = 0Hị: one or more of the restrictions under Ho does not holdThe sample size n =this test. The F-statistic is 2.15. What do you conclude? (R stands for reject and FTR standsfor fail to reject.)R Ho at a = 0.209000. The computer computes the heteroskedasticity-robust F-statistic forA)R Ho at a = 0.15R Ho at a = 0.10R Ho at a == 0.05R Ho at a =В)R Ho at a = 0.20R Ho at a = 0.15R Ho at a = 0.10R Ho at a == 0.05FTR Ho ata = 0.01%DC)R Ho at a =R Ho at a = 0.15R Ho at a = 0.10FTR Ho atFTR Ho ata = 0.05FTR Ho ata = 0.01FTR Ho at%3DD)R Ho at a = 0.20R Ho at a = 0.15FTR Ho ata = 0.10= 0.05a = 0.01%3DE)R Ho at a = 0.20FTR Ho atFTR Ho atFTR Ho atFTR Ho at= 0.15= 0.10a = 0.05= 0.01a =%3DF)FTR Ho atFTR Ho atFTR Ho atFTR Ho atFTR Ho ata = 0.20a = 0.15a = 0.100.05a = 0.01%3|2.

Y is regressed against X1, X2, X3, X4, X5, X6, X7, and X8. Consider the following test: Họ: B1 = B2 = B3 = B4 = Bs = 0 Hj: one or more of the restrictions under Họ does not hold The sample size n = 9000. The computer computes the heteroskedasticity-robust F-statistic for this test. The F-statistic is 2.15. What do you conclude? (R stands for reject and FTR stands for fail to reject.) R Ho at a = 0.20 R Ho at a = 0.15 R Họ at a = 0.10 R Ho at a = 0.05 R Ho at a = 0.01 R Ho at a = 0.20 R Ho at a = 0.15 R Ho at a = 0.10 R Ho at a = 0.05 FTR Ho at a = 0.01

Y is regressed against X1, X2, X3, X4, X5, X6, X7, and X8. Consider the following test: Họ: B1 = B2 = B3 = B4 = Bs = 0 Hj: one or more of the restrictions under Họ does not hold The sample size n = 9000. The computer computes the heteroskedasticity-robust F-statistic for this test. The F-statistic is 2.15. What do you conclude? (R stands for reject and FTR stands for fail to reject.) R Ho at a = 0.20 R Ho at a = 0.15 R Họ at a = 0.10 R Ho at a = 0.05 R Ho at a = 0.01 R Ho at a = 0.20 R Ho at a = 0.15 R Ho at a = 0.10 R Ho at a = 0.05 FTR Ho at a = 0.01

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

See similar textbooks

Similar questions

The following table shows the results of a two-factor analysis of variance with 2 levels of factor A, 3 levels of factor B, and a separate sample of n = 5 participants in each of the treatment conditions. Note that several values are missing in the table. What is the missing value for the F-ratio for the AxB interaction? Source SS df MS Between 80 xx A 8 xx xx F = xx B xx xx 20 F = xx AxB xx xx xx F = xx Within xx xx xx Total 176 xx Question options: a. 4 b. 2 c. 16 d. 8
60 illiterate preschool children were randomly divided into four groups of 15 children. Each group used a different method to teach reading skills. Afterwards, reading skills were tested with the following results.Group X ̄i group Xi S2i 1 65 224 2 62 241 3 68 233 4 61 245 Test the null hypothesis that the methods are equally effective. Use a five percent significance level. Functions from excel or R should not be used to test this.
A randomized block design was used to compare the means of three treatments within six blocks. Construct an ANOVA table showing the sources of variation and their respective degrees of freedom. Suppose that the analysis of variance calculations are SST 11.4, SSB 17.1, and Total SS 42.7. Complete the ANOVA table, showing all sums of squares, mean squares, and pertinent F-values.
Is the following statment true or false? Why? The p-value was obtained using the t-distribution table. The sample sizes of the two groups were 33 and 22, and the degree of freedom was 53. To find the p-value, it is necessary to look at the row corresponding to 53 at a significance level of 0.05 and two tails. The value of the critical test statistic is 2. The computed t-statistic, 0.2112, falls between 0 and 0.679 from the same row. The next step is to look at the alpha levels at the top of the table with two tails, corresponding to 0 and 0.679. These alpha levels are 1 and 0.5. Therefore, the p-value of the test, which is 0.75, will be between 1 and 0.5. The p-value was confirmed using the TI-84 obtaining a similar p-value of 0.7994. Traditional Students (G1) Online Students (G2) mean 2.97 2.95 standard deviation 0.34 0.35 sample size 33 22 significance level: 0.05 degree of freedom: 53
A transect is an archaeological study area that is 1/5 mile wide and 1 mile long. A site in a transect is the location of a significant archaeological find. Let x represent the number of sites per transect. In a section of Chaco Canyon, a large number of transects showed that x has a population variance σ2 = 42.3. In a different section of Chaco Canyon, a random sample of 24 transects gave a sample variance s2 = 50.3 for the number of sites per transect. Use a 5% level of significance to test the claim that the variance in the new section is greater than 42.3. Find a 95% confidence interval for the population variance. (a) What is the level of significance? (b) Find the value of the chi-square statistic for the sample. (Round your answer to two decimal places.) (C) Find the requested confidence interval for the population variance. (Round your answers to two decimal places.) Interpret the results in the context of the application.
A researcher is interested in determining whether background music has had an effect on participants' word recall. An experiment was conducted during which one group was given a list of words to study and then asked to recall the words while light jazz was playing in the background the entire time. The second group did the same thing in a completely silent environment. Below are percentages of correctly recalled words for each group.Conduct a two-samples t-test (equal variances assumed) to determiine if there was a significant difference between average percentages of recalled words in both groups.Assume alpha =0.05.In your answer, state the null and the research hypotheses and explain whether the results of the two-sample test led you to reject or fail to reject the null hypothesis, and what it means.Copy the data below into Google Sheets and perform t-test: two-sample (assuming equal variances). Copy/paste the results into the question response window. In your answer, state the…
A transect is an archaeological study area that is 1/5 mile wide and 1 mile long. A site in a transect is the location of a significant archaeological find. Let x represent the number of sites per transect. In a section of Chaco Canyon, a large number of transects showed that x has a population variance σ2 = 42.3. In a different section of Chaco Canyon, a random sample of 19 transects gave a sample variance s2 = 49.3 for the number of sites per transect. Use a 5% level of significance to test the claim that the variance in the new section is greater than 42.3. Find a 95% confidence interval for the population variance. (a) What is the level of significance?State the null and alternate hypotheses. Ho: σ2 = 42.3; H1: σ2 > 42.3Ho: σ2 > 42.3; H1: σ2 = 42.3 Ho: σ2 = 42.3; H1: σ2 < 42.3Ho: σ2 = 42.3; H1: σ2 ≠ 42.3 (b) Find the value of the chi-square statistic for the sample. (Round your answer to two decimal places.)What are the degrees of freedom?What assumptions are you making…
One of the two fire stations in a certain town responds to calls in the northern half of the town, and the other fire station responds to calls in the southern half of the town. The following is a list of response times (in minutes) for both of the fire stations (this data will be used for several problems). Both samples may be regarded as simple random samples from approximately normal populations so that the t- procedures are safe to use. Northern: 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 6, 7, 7, 8, 8, 8, 8, 8, 9, 9, 10, 10, 10, 10, 12 Sum = 192 Sum of Squared Deviations = 197.2 Southern: 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 6, 6, 7, 7, 8, 8, 8, 8, 8, 8, 9, 10, 10, 10, 12, 12, 12, 12, 13 Sum = 225 Sum of Squared Deviations = 231.5 Find and interpret a 95% confidence interval for the mean response time of the fire station that responds to calls in the northern part of town. Fill in blank 1 to report the bounds of the 95% Cl. Enter your answers as lower bound,upper bound with no…
An experiment has been conducted for four treatments with eight blocks. Complete the following analysis of variance table.
Find Variance within the groups, and variance between the groups, and the critical value
A major oil company has developed a new gasoline additive that is supposed to increase mileage. To test this hypothesis, ten cars are randomly selected. The cars are driven both with and without the additive. The results are displayed in the following table. Can it be concluded, from the data, that the gasoline additive does significantly increase mileage? Let d=(gas mileage with additive)−(gas mileage without additive). Use a significance level of α=0.05 for the test. Assume that the gas mileages are normally distributed for the population of all cars both with and without the additive. Car 1 2 3 4 5 6 7 8 9 10 Without additive 27.2 13.7 12.5 13.4 14.7 25.5 10.3 13.9 14.5 25 With additive 29.7 15.7 14.1 16.4 17.8 27.8 12.1 15 17.3 28 Step 2 of 5: Find the value of the standard deviation of the paired differences. Round your answer to two decimal places.
Random samples of resting heart rates are taken from two groups. Population 1 exercises regularly, and Population 2 does not. The data from these two samples is given below: Population 1: 66, 71, 66, 64, 63, 69, 70 Population 2: 77, 74, 69, 73, 74, 78, 68, 68 Is there evidence, at an a = 0.065 level of significance, to conclude that there those who exercise regularly have lower resting heart rates? Carry out an appropriate hypothesis test, filling in the information requested. A. The value of the standardized test statistic: -4.87322 B. The p-value is .00001 C. Your decision for the hypothesis test: A. Do Not Reject H₁. B. Do Not Reject Ho. C. Reject H₁. OD. Reject Ho.