In: Computer Science
1)
Confidence interval(CI) for normal distribution is (-z*/, +z*/)
here =101.82
n=6, =1.2
and Z value will be 1.96 which corresponds to (100-95)/2 %= 2.5%
z*/= 0.96
Therefore CI will be (101.82-0.96,101.82+0.96)= (100.86,102.78)
2)
Here we will use a t-distribution because n<30 and population mean is given as well .
Confidence interval(CI) for t-distribution is (-t*s/, +t*s/)
here =240
n=10, s=25
df=n-1 =10-1=9
By looking up in t-distribution table we find the value for df=9 and 2.5% value we get t=2.262
t*s/= 17.88
Therefore CI will be (240-17.88,240+17.88)= (222.12,257.88)
3)
CI for population proportion is( , )
n=510
Here alpha=1-0.9 =0.1
Z value for alpha/2 (.1/2=.05) is 0.13 ( By looking in the z table)
=57/510=0.112
1-=1-.112=0.888
=0.00182
Therefore CI will be (0.112-0.00182,0.112+0.00182)=(0.11018,0.11382)
4)
CI for difference in proportions is
( , )
n1=100,n2=75
=0.65
=0.33
= 1-0.65=0.35
=1-.33=0.67
Since it is the same 90% confidence interval as in previous sum z value will be 0.13
=0.0098
=0.65-0.33=0.32
Therefore CI will be (0.32-0.0098,0.32+0.0098)=(31.9902,32.0098)