In: Statistics and Probability
Solution-A:
For the given sample ,
n=12
df=n-1=12-1=11
sample standaed dev=s=10.69799
alpha/2=0.05/2=0.025
1-alpha/2=1-0.025=0.975
chi sq alpha/2==CHISQ.INV.RT(0.025,11)=21.92004926
chi sq 1-alpha/2==CHISQ.INV.RT(0.975,11)=3.815748252
95% confidence interval for sigma
sqrt(n-1)*s^2/chisq(1-alpha/2)<sigma<sqrt(n-1)*s^2/chisq(alpha/2)
sqrt((12-1)*10.69799^2/21.92004926)<sigma<sqrt((12-1)*10.69799^2/3.815748252)
7.578404<sigma< 18.16388
7.578404<sigma< 18.16388
There is 95% confidence that the population SD is between 7.578 and 18.164
Solution-b:
Rcode:
ph <- c(5.20, 5.72, 4.38, 4.80, 5.02, 4.58, 4.74, 5.19, 5.34,
4.76, 4.56, 5.68)
t.test(ph,conf.level = 0.95)
output:
data: ph
t = 39.693, df = 11, p-value = 3.147e-13
alternative hypothesis: true mean is not equal to 0
95 percent confidence interval:
4.720388 5.274612
sample estimates:
mean of x
4.9975
ANSWER;
95 percent confidence interval:
4.720388 5.274612
there is 95% confidence that the pop mean of ph of rain water is between 4.720388 and 5.274612
Solution-c:
ph <- c(5.20, 5.72, 4.38, 4.80, 5.02, 4.58, 4.74, 5.19, 5.34,
4.76, 4.56, 5.68)
t.test(ph,conf.level = 0.99)
One Sample t-test
data: ph
t = 39.693, df = 11, p-value = 3.147e-13
alternative hypothesis: true mean is not equal to 0
99 percent confidence interval:
4.606468 5.388532
sample estimates:
mean of x
4.9975
ANSWER::
99 percent confidence interval:
4.606468 and 5.388532
a. there is 99% confidence that the population mean ph of rain water is between 4.606468 and 5.388532