MATLAB: An Introduction with Applications
6th Edition
ISBN: 9781119256830
Author: Amos Gilat
Publisher: John Wiley & Sons Inc
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What is the negative critical value of a two-tailed One-Sample Z-Test for
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- The average annual miles driven per vehicle in the United States is 11.1 thousand miles, with o = 600 miles. Suppose that a random sample of 31 vehicles owned by residents of Chicago showed that the average mileage driven last year was 10.8 thousand miles. Does this indicate that the average miles driven per vehicle in Chicago is different from (higher or lower than) the national average? Use a 0.05 level of significance. What are we testing in this problem? O single mean O single proportion (a) What is the level of significance? State the null and alternate hypotheses. о Но: р3D 11.1; Hі: р> 11.1 O Ho: H = 11.1; H1: µ > 11.1 O Ho: µ = 11.1; H1: µ # 11.1 о Hо: р 3 11.1; Hi: р# 11.1 O Ho: p = 11.1; H1: p 0.500 O 0.250 < P-value < 0.500 O 0.100 < P-value < 0.250 O 0.050 < P-value < 0.100 O 0.010 < P-value < 0.050 O P-value < 0.010arrow_forwardGoogle Sheets can be used to find the critical value for an F distribution for a given significance level. For example, to find the critical value (the value above which you would reject the null hypothesis) for α=0.05 with dfbetween=4 and dfwithin=30, enter =FINV(0.05,4,30) Enter this into Google Sheets to confirm you obtain the value 2.69. You conduct a one-factor ANOVA with 8 groups and 10 subjects in each group (a balanced design). Use Google Sheets to find the critical values for α=0.1 and α=0.02 (report accurate to 3 decimal places).F0.1=F0.02=arrow_forwardA certain breed of rat shows a mean weight gain of 65g during the first 3 months of life. A random sample of 34 rats of this breed are fed a new diet from birth until age 3 months. These 34 rats have a mean weight gain of 60.75g and a standard deviation of 3.84g. We are interested in testing whether there is reason to believe, at the 0.05 significance level, that the new diet is causing a change in the average amount of weight gained in this breed of rats. By comparing the test statistic and the critical value, what conclusion can we draw at the 0.05 significance level? Since the test statistic is more extreme than the critical value, we reject the null hypothesis. We have statistically significant evidence to conclude that average weight gain during the first 3 months of a certain breed of rat's life on the new diet is not equal to 65g. Since the test statistic is NOT more extreme than the critical value, we fail to reject the null hypothesis. We do not have statistically significant…arrow_forward
- What is the critical value of a one-tailed greater than One-Sample Z-Test for Mean where alpha equals 0.05?arrow_forwardFor an upper-tail test at the 1% significance level, what is the appropriate critical value when a sample of size 28 is drawn from a normal population with unknown standard deviation?arrow_forwardWe want to know if there is a difference between the mean list price of a three bedroom home, μ3, and the mean list price of a four bedroom home, μ4, H0:μ3=μ4 versus Ha:μ3≠μ4. Suppose we get a p-value of 0.1325, which is the proper conclusion of this test? a) We accept the null hypothesis at 5% significance. There is extremely strong evidence of a significance difference between the mean list price of three bedroom homes and the mean list price of four bedroom homes. b) We reject the null hypothesis at 5% significance. There is no evidence of a significance difference between the mean list price of three bedroom homes and the mean list price of four bedroom homes. c) We fail to reject the null hypothesis at 5% significance. There is no evidence of a significance difference between the mean list price of three bedroom homes and the mean list price of four bedroom homes. d) We reject the null hypothesis at 5% significance. There is extremely strong evidence of a significance difference…arrow_forward
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