Statistics for Business & Economics
16th Edition
ISBN: 9781337325448
Author: Anderson
Publisher: CENGAGE C
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Textbook Question
Chapter 11.2, Problem 14E
A sample of 16 items from population 1 has a sample variance
- a. What is your conclusion using the p-value approach?
- b. Repeat the test using the critical value approach.
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For a hypothesis test of H0:μ=μ0H0:μ=μ0 against the alternative Ha:μ<μ0Ha:μ<μ0, the z-test statistics is found to be 2.00. This finding is
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Chapter 11 Solutions
Statistics for Business & Economics
Ch. 11.1 - Find the following chi-square distribution values...Ch. 11.1 - A sample of 20 items provides a sample standard...Ch. 11.1 - A sample of 16 items provides a sample standard...Ch. 11.1 - The variance in drug weights is critical in the...Ch. 11.1 - John Calipari, head basketball coach for the 2012...Ch. 11.1 - Americans spend nearly 7 billion on Halloween...Ch. 11.1 - To analyze the risk, or volatility, associated...Ch. 11.1 - Consider a day when the Dow Jones industrial...Ch. 11.1 - An automotive part must be machined to close...Ch. 11.1 - Consumer Reports uses a 100-point customer...
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- A hypothesis test was conducted, at α = 0.05, to determine whether a certain chemical compound lasts longer than 30 seconds under a certain specified condition. The hypotheses used were: H0: µ = 30 Ha: µ > 30 A sample mean of 37.4 seconds was obtained from a sample of size n = 80. All statistical assumptions were met, and a p-value of p = 0.0089 was obtained. Which of the following is correct? a) If the null hypothesis were in reality true that the population mean was equal to 30, then the probability of observing a sample mean of 37.4 seconds from a sample of size n = 80 would be only .0089. b) If the null hypothesis were in reality false that the population mean was equal to 30, then the probability of observing a sample mean of 37.4 seconds (or less) from a sample of size n = 80 would be only .0089. c) If the null hypothesis were in reality true that the population mean was equal to 30, then the probability of observing a sample mean of 37.4 seconds (or greater) from…arrow_forwardIf the critical t is ±1.796 and the obtained t is -2.09, what decision would you make regarding the null hypothesis?arrow_forwarda hypothesis test produces a t statistic of t=2.3. if the researcher is using a two tailed test with a=0.05 how large does the sample have to bw in order to reject the null hypothesis?arrow_forward
- Q1. An anthropologist wants to collect data to determine whether the two different cultural groups that occupy an isolated Pacific Island grow to be different heights. The results of his samples of the heights of adult females are as follows Do these samples constitute enough evidence to reject the null hypothesis that the heights of the two groups the same? Set alpha to .05.arrow_forwardBased on the following values: Mathematical expectation: E(X) = ∑Xn1 = 1204 = 30 E(Y) = ∑Yn2 = 1884 = 47 E(Z) = ∑Zn3 = 1784 = 44.5 Variance: var(X) = 1n1-1∑X2-1n∑X2 = 14-13678-12024 =26 var(Y) = 1n2-1∑Y2-1n∑Y2 = 14-18890-18824 =18 var(Z) = 1n3-1∑Z2-1n3∑Z2 = 14-18322-17824 =133.6667 Root mean square deviation sd(X)= var(X) =26 = 5.0990 sd(Y)= var(Y) =18 = 4.2426 sd(Z)= var(Z) =133.6667 = 11.5614 Correlation coefficients X Y Z 25 44 28 28 43 48 37 49 55 30 52 47 1. Expectation 120 188 178 2. Variance 26 18 133.6667 3. root mean square deviation 5.099 4.2426 11.5614 Check the hypotheses in the image below, in accordance with the tablic values.arrow_forwardBased on the following values: Mathematical expectation: E(X) = ∑Xn1 = 1204 = 30 E(Y) = ∑Yn2 = 1884 = 47 E(Z) = ∑Zn3 = 1784 = 44.5 Variance: var(X) = 1n1-1∑X2-1n∑X2 = 14-13678-12024 =26 var(Y) = 1n2-1∑Y2-1n∑Y2 = 14-18890-18824 =18 var(Z) = 1n3-1∑Z2-1n3∑Z2 = 14-18322-17824 =133.6667 Root mean square deviation sd(X)= var(X) =26 = 5.0990 sd(Y)= var(Y) =18 = 4.2426 sd(Z)= var(Z) =133.6667 = 11.5614 Correlation coefficients X Y Z 25 44 28 28 43 48 37 49 55 30 52 47 1. Expectation 120 188 178 2. Variance 26 18 133.6667 3. root mean square deviation 5.099 4.2426 11.5614 Check the hypothesis in the image below. The table values are also therearrow_forward
- Based on the following values: Mathematical expectation: E(X) = ∑Xn1 = 1204 = 30 E(Y) = ∑Yn2 = 1884 = 47 E(Z) = ∑Zn3 = 1784 = 44.5 Variance: var(X) = 1n1-1∑X2-1n∑X2 = 14-13678-12024 =26 var(Y) = 1n2-1∑Y2-1n∑Y2 = 14-18890-18824 =18 var(Z) = 1n3-1∑Z2-1n3∑Z2 = 14-18322-17824 =133.6667 Root mean square deviation sd(X)= var(X) =26 = 5.0990 sd(Y)= var(Y) =18 = 4.2426 sd(Z)= var(Z) =133.6667 = 11.5614 Correlation coefficients X Y Z 25 44 28 28 43 48 37 49 55 30 52 47 1. Expectation 120 188 178 2. Variance 26 18 133.6667 3. root mean square deviation 5.099 4.2426 11.5614 Check the hypothesis in the images. The table values are in the other image belowarrow_forwardBased on the following values: Mathematical expectation: E(X) = ∑Xn1 = 1204 = 30 E(Y) = ∑Yn2 = 1884 = 47 E(Z) = ∑Zn3 = 1784 = 44.5 Variance: var(X) = 1n1-1∑X2-1n∑X2 = 14-13678-12024 =26 var(Y) = 1n2-1∑Y2-1n∑Y2 = 14-18890-18824 =18 var(Z) = 1n3-1∑Z2-1n3∑Z2 = 14-18322-17824 =133.6667 Root mean square deviation sd(X)= var(X) =26 = 5.0990 sd(Y)= var(Y) =18 = 4.2426 sd(Z)= var(Z) =133.6667 = 11.5614 Correlation coefficients X Y Z 25 44 28 28 43 48 37 49 55 30 52 47 1. Expectation 120 188 178 2. Variance 26 18 133.6667 3. root mean square deviation 5.099 4.2426 11.5614 Check the hypothesis in the image below. The tablic values are also therearrow_forwardConsider the following null and alternative hypotheses for testing whether the average GPA of university students is different than 3.2. H0:u = 3.2 Ha:u not= 3.2 In this case, the consequence of Type I error is... a. assuming university students have a GPA equal to 3.2, when in fact it is not equal to 3.2. b. assuming university students have a lower GPA than 3.2, when in fact it is equal to 3.2. c. assuming university students have a higher GPA than 3.2, when in fact it is equal to 3.2. d. assuming university students have a GPA not equal to 3.2, when in fact it is equal to 3.2.arrow_forward
- A hypothesis test produces a t statistic of t = +2.19. If the researcher is conducting a two-tailed hypothesis test with α = .05, how large does the sample have to be in order to reject the null hypothesis?arrow_forwardSuppose we are testing the null hypothesis H_0 : \mu = 32H0:μ=32 and the alternative hypothesis H_a : \mu \ne 32Ha:μ=32, for a normal population with population standard deviation of \sigma =σ= 3.5. A random sample of sixteen observations are drawn from the population, and we find the sample mean of these observations is \overline{x} =x= 30.5. Compute the P-value for this test. Give your answer to four decimal places.arrow_forwardWhich of the following statement is CORRECT? A. If the null hpothesis is false, the sample means are considered the same. B. μ(mu)1 will never be equal to μ(mu)2 C. If the null hypothesis is true, there is a difference between population means. D. If the null hypothesis is true, there is no difference between the population means.arrow_forward
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