Looking at the table below what do you conclude in a test of weekly expenditure on fruit between females and non-females if one were testing the null hypothesis of no difference against a 1-s alternative that females spend more on fruit at the 5% significance level: Two-sample t test with equal variances 13.812 15.510 Group Obs Mean Std. err. Std. dev. [95% conf. interval] Female 377 14.661 .4317 Non-Female 206 13.801 Combined 583 14.357 diff .8602 12.793 14.809 8.381 .5113 7.339 6956 .6957 8.033 13.703 15.011 -.5061 2.2265 diff = mean(Female) - mean(Non-Female) HO: diff = 0 Ha: diff < 0 Pr(T |t|) = 0.217 O With a p-value of 0.892 we do not reiect HO t = 1.237 Degrees of freedom = 581 Ha: diff > 0 Pr(T> t) = 0.108
Looking at the table below what do you conclude in a test of weekly expenditure on fruit between females and non-females if one were testing the null hypothesis of no difference against a 1-s alternative that females spend more on fruit at the 5% significance level: Two-sample t test with equal variances 13.812 15.510 Group Obs Mean Std. err. Std. dev. [95% conf. interval] Female 377 14.661 .4317 Non-Female 206 13.801 Combined 583 14.357 diff .8602 12.793 14.809 8.381 .5113 7.339 6956 .6957 8.033 13.703 15.011 -.5061 2.2265 diff = mean(Female) - mean(Non-Female) HO: diff = 0 Ha: diff < 0 Pr(T |t|) = 0.217 O With a p-value of 0.892 we do not reiect HO t = 1.237 Degrees of freedom = 581 Ha: diff > 0 Pr(T> t) = 0.108
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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![Looking at the table below what do you conclude in a test of weekly expenditure on fruit between females and non-females if one were testing the null hypothesis of no difference against a 1-sided
alternative that females spend more on fruit at the 5% significance level:
Two-sample t test with equal variances
Group Obs Mean Std. err. Std. dev. [95% conf. interval]
Female 377 14.661 .4317 8.381 13.812 15.510
Non-Female 206 13.801 .5113 7.339 12.793 14.809
Combined 583 14.357 .6956 8.033 13.703 15.011
diff
.8602 .6957
-.5061 2.2265
diff= mean(Female) - mean(Non-Female)
HO: diff = 0
Ha: diff < 0
Pr(Tt) = 0.892
Ha: diff != 0
Pr(|T| > |t|) = 0.217
With a p-value of 0.892 we do not reject HO
With a p-value of 0.892 we accept HO
With a p-value of 0.108 we do not reject HO
With a p-value of 0.217 we do not reject HO
t = 1.237
Degrees of freedom = 581
Ha: diff > 0
Pr(T>t) = 0.108](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3c13ca9d-51c9-4c63-be3d-a7cc26f8ce3f%2F7f936279-6441-4df1-87ee-e1397b4535e6%2F880q7cp_processed.png&w=3840&q=75)
Transcribed Image Text:Looking at the table below what do you conclude in a test of weekly expenditure on fruit between females and non-females if one were testing the null hypothesis of no difference against a 1-sided
alternative that females spend more on fruit at the 5% significance level:
Two-sample t test with equal variances
Group Obs Mean Std. err. Std. dev. [95% conf. interval]
Female 377 14.661 .4317 8.381 13.812 15.510
Non-Female 206 13.801 .5113 7.339 12.793 14.809
Combined 583 14.357 .6956 8.033 13.703 15.011
diff
.8602 .6957
-.5061 2.2265
diff= mean(Female) - mean(Non-Female)
HO: diff = 0
Ha: diff < 0
Pr(Tt) = 0.892
Ha: diff != 0
Pr(|T| > |t|) = 0.217
With a p-value of 0.892 we do not reject HO
With a p-value of 0.892 we accept HO
With a p-value of 0.108 we do not reject HO
With a p-value of 0.217 we do not reject HO
t = 1.237
Degrees of freedom = 581
Ha: diff > 0
Pr(T>t) = 0.108
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