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. ttest mpg==19 One-sample t test Variable Obs Mean Std. err. Std. dev. [95% conf. interval] mpg 16 18.15625 .371031 1.484124 17.36542 18.94708 t = -2.2741 mean = mean(mpg) HO: mean = 19 Degrees of freedom = 15 Ha: mean < 19 Ha: mean != 19 Ha: mean > 19 Pr(T < t) = 0.0190 Pr(|T| > [t|) = 0.0381 Pr(T > t) = 0.9810 A. The sample mean is 19 mpg. B. The sample standard deviation is 0.371. C. The two-sided null hypothesis is rejected at the 95% confidence lev D. There is a 0.981 probability that the fuel economy of this model is above 19 mpg.

. ttest mpg==19 One-sample t test Variable Obs Mean Std. err. Std. dev. [95% conf. interval] mpg 16 18.15625 .371031 1.484124 17.36542 18.94708 t = -2.2741 mean = mean(mpg) HO: mean = 19 Degrees of freedom = 15 Ha: mean < 19 Ha: mean != 19 Ha: mean > 19 Pr(T < t) = 0.0190 Pr(|T| > [t|) = 0.0381 Pr(T > t) = 0.9810 A. The sample mean is 19 mpg. B. The sample standard deviation is 0.371. C. The two-sided null hypothesis is rejected at the 95% confidence lev D. There is a 0.981 probability that the fuel economy of this model is above 19 mpg.

MATLAB: An Introduction with Applications
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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ttest mpg==19 One-sample t test Variable Obs Mean Std. err. Std. dev. [95% conf. interval] mpg 16 18.15625 .371031 1.484124 17.36542 18.94708 mean = mean(mpg) HO: mean = 19 t = -2.2741 Degrees of freedom = 15 Ha: mean < 19 Ha: mean != 19 Ha: mean > 19 Pr(T < t) = 0.0190 Pr(|T| > |t|) = 0.0381 Pr(T > t) = 0.9810 A. The sample mean is 19 mpg. B. The sample standard deviation is 0.371. C. The two-sided null hypothesis is rejected at the 95% confidence level. D. There is a 0.981 probability that the fuel economy of this model is above 19 mpg.
Transcribed Image Text:ttest mpg==19 One-sample t test Variable Obs Mean Std. err. Std. dev. [95% conf. interval] mpg 16 18.15625 .371031 1.484124 17.36542 18.94708 mean = mean(mpg) HO: mean = 19 t = -2.2741 Degrees of freedom = 15 Ha: mean < 19 Ha: mean != 19 Ha: mean > 19 Pr(T < t) = 0.0190 Pr(|T| > |t|) = 0.0381 Pr(T > t) = 0.9810 A. The sample mean is 19 mpg. B. The sample standard deviation is 0.371. C. The two-sided null hypothesis is rejected at the 95% confidence level. D. There is a 0.981 probability that the fuel economy of this model is above 19 mpg.
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