Assignment 7
1. So that we will know where 50% of his subjects are located (which is below the median), and the other 50% of his subjects are located (which is above the median).
2.
> y <- rnorm(16, 0,2)
> hist(y)
z=X-μσn=-0.4088-0.5216=-1.818
3.
> x <- rnorm(25,0,2)
> hist(x)
> mean(x)
[1] -0.660643
t=X-μsdn=-0.660643-0.5225≈-2.90
4. Number of degrees of freedom = 25 – 1 = 24
5.
t.test(x,alternative="two.sided",mu=0.25)
One Sample t-test
data: x
t = -2.5001, df = 24, p-value = 0.01965
alternative hypothesis: true mean is not equal to 0.25
95 percent confidence interval:
-1.41240336 0.09111734
sample estimates:
mean of x
-0.660643
The p-value is 0.01965 < 0.05 => null hypothesis is rejected.
6. The confidence interval means that there is a 95% chance that the true value is within the interval.
7. > wilcox.test(x,mu=0.2)
Wilcoxon signed rank test
data: x
V = 85, p-value = 0.03668
alternative hypothesis: true location is not equal to 0.2
0.03668 > 0.05 => null hypothesis is not rejected.
8.
Test statistics, z = 5-02.4≈2.08
9. I take 95% confidence limit to test for null hypothesis. So, probability of it being true is 0.95.
10. Sample size > 30.
11. The uniform distribution should be between -20 and 20 since the maximum is 40 units away from the minimum, and the null hypothesis is that the mean is zero. From probability to get anything from 0 to 20 in the uniform distribution is 50%. So, for 95% confidence limits, first divide 95 by 2 to get 95/2 = 47.5%. Then, ask what is the value V within 0 to 20, which will give us a 47.5% probability to get anything from 0 to V in this uniform distribution.
So, V = (47.5/50)(20-0) = 19
So, the 95% confidence limits is -19, 19.
12. It tends to become more like the normal distribution, and with a higher peak.
