Ask your own question, for FREE!
Mathematics 16 Online
OpenStudy (kitkat16):

Help!! If the null hypothesis states that there is no difference between two household's water consumption. The sample means difference is 8.5 liters and the standard deviation of the distribution of the difference in sample means is 4.5 liters. The null hypothesis will be accepted if we choose the 99.7% confidence interval. True or false? Anyone know how to figure this out? I want to say true because I thought null meant no difference and it was always 99.7

OpenStudy (anonymous):

The null hypothesis states there is no difference between the households' water consumption; the observed difference is 8.5 liters. Assuming the distribution of differences is normal with mean zero and standard deviation 4.5 liters (corresponding to the null hypothesis), the probability of observing such a difference is computed as follows... \[Z=\frac1\sigma(x-\mu)=\frac1{4.5}(8.5-0)=\frac{17}9\approx1.89\]

OpenStudy (anonymous):

now, what is the probability of observing an event this rare? $$P(|z|>Z)=P(z<-Z)+P(z>Z)=2\cdot P(z>1.89)\approx0.0589068$$

OpenStudy (kitkat16):

omgosh I have never seen anything like that yet

OpenStudy (anonymous):

so there is a 5.89% chance of observing a difference at least this significant with the null hypothesis holding; this is not within the 0.3% margin of significance corresponding to our confidence interval

OpenStudy (kitkat16):

is there a different way to figure this out because I thought i was suppose to divide something.

OpenStudy (anonymous):

you can also compute the confidence interval directly

OpenStudy (kitkat16):

my choices were 68% , 95% and 99.7%

OpenStudy (anonymous):

it says true or false, not sure how there are 3 choices?

OpenStudy (kitkat16):

Well I was asking is it true my reasons for picking were I thought null meant no difference and it was always 99.7

OpenStudy (anonymous):

99.7% confidence corresponds to a critical value \(z^*\approx2.96774\), giving a confidence interval of \(\mu\pm z^*\sigma\approx\pm2.96774\cdot4.5\approx\pm13.35483\); since \(-13.35483\le8.5\le13.35483\) it follows that we cannot reject the null hypothesis at 99.7% confidence

OpenStudy (kitkat16):

did I state my question wrong maybe. the null hypothesis must be accepted if we choose the ? % confidence interval.

OpenStudy (anonymous):

similarly, 95% confidence corresponds to a critical value \(z^*=1.95996\) giving a confidence interval of \(\pm8.81982\) so \(-8.81982\le8.5\le8.81982\) meaning we cannot reject the null hypothesis at 95% either

OpenStudy (anonymous):

the only confidence level for which we can reject the null hypothesis is 68%

OpenStudy (kitkat16):

so its 95%

OpenStudy (anonymous):

the 'shortcut' way is to use the empirical rule for which 68% corresponds for up to one standard deviation away (difference between -4.5 and 4.5 liters), whereas 95% corresponds to 2 standard deviations away (diff between -9 and 9 liters), and 99.7% to 3 standard deviations away (diff between -13.5 and 13.5 liters); only with the first confidence level is the observed difference more extreme, so both 95% and 99.7% allow one to 'accept' the null hypothesis

OpenStudy (kitkat16):

thank you

OpenStudy (kitkat16):

the null hypothesis must be accepted if we choose the 95 % confidence interval.

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!