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Question
A study was recently conducted to see whether the mean starting salaries for graduates of engineering, business, healthcare, and computer information systems majors differ. A random sample of 8 graduates was selected from each major. The following chart shows some of the results of the ANOVA computations; however, some of the output is missing. Given what is available, the proper conclusion to reach based on the sample data is that the population means could be equal using a 0.05 level of significance.ANOVA: Single Factor
Answer
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Related questions
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Six food critics each visited and rated four different restaurants. Each critic visited each restaurant on three separate occasions and recorded a score for each visit. The number of degrees of freedom in the interaction row of the ANOVA table is 15.
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Three brands of running shoes are each tested by 10 different runners. The amount of wear on the sole of the shoes is then measured. The objective is to determine if there is any difference among the three brands of shoes based on how long the soles last. The degrees of freedom for testing whether the brands of shoes differ are D1 = 9 and D2 = 2.
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A company has established an experiment with its production process in which three temperature settings are used and five elapsed times are used for each setting. The company then produces one product under each and measures the resulting strength of the product. This experimental design is called a randomized complete block design.
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Three brands of running shoes are each tested by 10 different runners. The amount of wear on the sole of the shoes is then measured. The objective is to determine if there is any difference among the three brands of shoes based on how long the soles last. The null hypothesis is:H0 : 1 = 2 = 3.
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Three brands of running shoes are each tested by 10 different runners. The amount of wear on the sole of the shoes is then measured. The objective is to determine if there is any difference among the three brands of shoes based on how long the soles last. This means that there is one factor with 10 levels and 3 blocks.
Q:
A randomized complete block analysis of variance allows the analyst to control for sources of variation that might adversely affect the analysis by using the concept of paired samples.
Q:
As a step in establishing its rates, an automobile insurance company is interested in determining whether there is a difference in the mean highway speeds that drivers of different age groups drive. To help answer this question, it has selected a random sample of drivers in three age categories: under 21, 21-50, and over 50. The engineers then recorded the drivers' speeds at a designated point on a highway in the state. The subjects were unaware that their speed was being recorded. The following one-way ANOVA output was generated from the sample data. Based upon this output, if the significance level is 0.05, the engineers should conclude that the mean speeds may all be equal since the p-value is less than alpha.
ANOVA: Single Factor
Q:
A chain of fast food restaurants wants to compare the average service times at three different restaurants. It wants to conduct a hypothesis test to determine if all three means are the same or not, at the 0.05 level of significance. If n = 7 observations are taken at each of the three restaurants, the critical value is F = 3.555.
Q:
A national car rental agency is interested in determining whether the mean days that customers rent cars is the same between three of its major cities. The following data reflect the number of days people rented a car for a sample of people in each of three cities. Assuming that a one-way analysis of variance is to be performed, the total sum of squares is computed to be approximately 120.9. Boston
Dallas
Seattle 5
7
7 3
7
5 7
11
8 1
5
11 2
7 3
Q:
In one-way analysis of variance, the within-sample variation is not affected by whether the null hypothesis is true or not.
Q:
In a one-way analysis of variance test, the following null and alternative hypotheses are appropriate:
H0 : μ1 = μ2 = μ3
Hα : μ1 ≠μ2 ≠μ3
Q:
In conducting one-way analysis of variance, the sample size for each group must be equal.
Q:
In a one-way analysis of variance design, there is a single factor of interest but there may be multiple levels of the factor.
Q:
A PC company uses two suppliers for rechargeable batteries for its notebook computers. Two factors are important quality features of the batteries: mean use time and variation. It is desirable that the mean use time be high and the variability be low. Recently, the PC maker conducted a test on batteries from the two suppliers. In the test, 9 randomly selected batteries from Supplier 1 were tested and 12 randomly selected batteries from Supplier 2 were tested. The following results were observed:Supplier 1 Supplier 2n1 = 9 n2 = 121 = 67.25 min 2 = 72.4 mins1 = 11.2 min s2 = 9.9 minBased on these sample results, can the PC maker conclude that a difference exists between the two batteries with respect to the population mean use time? Test using a 0.10 level of significance.
Q:
In performing a one-tailed test for the difference between two population variances, which of the following statements is true?
A) The level of alpha needs to be doubled before finding the F-critical value in the table.
B) The sample variance that is predicted to be larger in the alternative hypothesis goes in the numerator when forming the F-test statistic.
C) You always place the larger of the two sample variances in the numerator.
D) The alternative hypothesis must contain the equality.
Q:
The Russet Potato Company has been working on the development of a new potato seed that is hoped to be an improvement over the existing seed that is being used. Specifically, the company hopes that the new seed will result in less variability in individual potato length than the existing seed without reducing the mean length. To test whether this is the case, a sample of each seed is used to grow potatoes to maturity. The following information is given: Old Seed
New Seed Number of Seeds = 11
Number of Seeds = 16 Average length = 6.25 inches
Average length = 5.95 inches Standard Deviation = 1.0 inches
Standard Deviation = 0.80 inches The on these data, if the hypothesis test is conducted using a 0.05 level of significance, the calculated test statistic is:
A) = 1.25
B) = 0.80
C) = 0.64
D) = 1.56
Q:
A manufacturer of industrial plywood has a contract to supply a boat maker with a large amount of plywood. One of the specifications calls for the standard deviation in thickness to not exceed .03 inch. A sample of n = 30 sheets was sampled randomly from a recent production run. The mean thickness was right at the 3/4 inch target thickness and the sample standard deviation was .05 inch. Testing at the 0.05 level of significance, which of the following is true?A) The test statistic is approximately 80.56.B) The critical value is approximately x2 = 43.773.C) The test statistic is approximately 48.333.D) Based on the sample data, there is no evidence to suggest that the plywood is not meeting the specifications.
Q:
If a hypothesis test for a single population variance is to be conducted, which of the following statements is true?
A) The null hypothesis must be stated in terms of the population variance.
B) The chi-square distribution is used.
C) If the sample size is increased, the critical value is also increased for a given level of statistical significance.
D) All of the above are true.
Q:
When conducting a one-tailed hypothesis test of a population variance using a sample size of n = 24 and a 0.10 level of significance, the critical value is:
A) 32.0069
B) 35.1725
C) 33.1962
D) 36.4150
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An analyst plans to test whether the standard deviation for the time it takes bank tellers to provide service to customers exceeds the standard of 1.5 minutes. The correct null and alternative hypotheses for this test are:
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A frozen food company that makes burritos currently has employees making burritos by hand. It is considering purchasing equipment to automate the process and wants to determine if the automated process would result in lower variability of burrito weights. It takes a random sample from each process as shown below. Process 1 (by hand)
Process 2 (automated) n = 10
n = 8 s = 0.2 ounces
s = 0.17 ounces To conduct a hypothesis test using a 0.05 level of significance, the critical value is 3.347.
Q:
In a two-tailed hypothesis test involving two population variances, if the null hypothesis is true then the F-test statistic should be approximately equal to 1.0.
Q:
One of the major automobile makers has developed two new engines. At question is whether the two engines have the same variability with respect to miles per gallon. To test this, the following information is available: Engine 1Engine 2n1 = 7n2 = 91 = 28.72 = 33.4s1 = 3.4s2 = 4.12Based on this situation and the information provided, the appropriate null and alternative hypotheses are:H0 : Ha : =
Q:
The managers for a vegetable canning facility claim the standard deviation for the ounces per can on the new automated line is less than for the older manual line. Given this, the correct null and alternative hypotheses for performing the statistical test are:H0 : 1 = 2Ha : 1 2
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A potato chip manufacturer has two packaging lines and wants to determine if the variances differ between the two lines. They take a sample of n= 15 bags from each line and find the following: The value of the test statistic is F = 1.5
Q:
In a hypothesis test involving two population variances, if the null hypothesis states that the two variances are strictly equal, then the test statistic is a chi-square statistic.
Q:
When using a chi-square test for the variance of one population, we are assuming that the population is normally distributed.
Q:
For a given significance level, increasing the sample size will tend to increase the chi-square critical value used in testing the null hypothesis about a population variance.
Q:
Assume a sample of size n = 12 has been collected. To perform a hypothesis test of a population variance using a 0.05 level of significance, where the null hypothesis is:H0: 2 = 25The upper tail critical value is 21.92.