Question 1081012: Can someone please help me check if my hw is correct?
ME=.135
phat=.6
We gather a sample of 50 people from a gym. the same prop. of women is .6.
Test the hypothesis that U is 55% using significance lvl of a=.1
i.What is the null and alternative hypothesis?
null-u=55 alternative-u does not equal 55
ii.Is the hypothesis considered one sided or two sided?
two sided
iii.If the hypothesis is true, what is the distribution of X(bar)?
60%
iv.Is the answer to part (iii) the same as 1c?
yes
v.What is the value of the ZTest
1.645
vi.What does the ZTest measure
verify if the means are different when variance and sample sizes are known
vii.Is the hypothesis likely or not likely to be true?
likely to be true
viii.What is ZCritical
1.96
ix.If your conclusion in part viii is wrong, what type of error did you make?
type I
x.In this example, what is the probability of a type I error
??
xi.What is the P-value
.042
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! The standard error is sqrt(.6*.4/50)=0.0693
It is a two-sided test, and the z value is 1.645
The distribution of the mean is (0.60) with SE 0.0693 and 95% CI interval is +/- 0.135
Note, for 90% CI the interval is +/- 0.1140
I am not clear on some of the questions, such as verify if the means are different when variance and sample sizes are known.
At the 90% level, the confidence interval is (0.486, 0.714) and it is even wider at the 95% level. The CI contains the parameter 0.55 or 55%, so fail to reject Ho. If Ho should have been rejected, but we accepted a false null hypothesis, that is a Type II error. Here, we fail to reject Ho at both the 5 and 10% levels, and the p-value for this is 0.48.
Note: the z-test starts with the assumption that there is no difference. The z-test for a two way test with 0.10 significance is >|1.645| The critical values are set which, if exceeded, say that the value that was found was so not likely to be due to chance that the null hypothesis should be rejected. The p-value is the likelihood that if the null hypothesis were true, the likelihood of finding a result or greater (in either direction away from Ho) is in this instance 0.48. Low p-values say that the change is highly unlikely to be due to random factors.
Also note that we are looking for a difference of a proportion from a hypothesized proportion. It happens to be the case that the mean and the variance in a proportion test are not independent, which means if you know the mean, you also know the variance, unlike a Gaussian or normal distribution.
|
|
|
