1 A survey of 100 businesses revealed that the mean aftertax
1) A survey of 100 businesses revealed that the mean after-tax profit was $80,000 and the standard deviation was $12,000.
Determine the 90% confidence interval estimate of the mean after-tax profit.
[78032, 81968]
[76130, 83870]
[77540, 82460]
[76904, 83096]
[77060, 82940]
2)
It is desired to estimate the mean tensile strength for roof hangers.
It is known that the standard deviation of measurements of tensile strength is 0.25. (Units are Newton per square meter.)
As it is very important for safety, the 99% confidence interval needs to have a margin smaller than 0.06.
What is the minimum required sample size?
85
116
261
16
e. 463
3) One is interested in the percentage of doctors who recommend aspirin rather than other painkillers for their patients with headaches.
A random sample of 145 doctors\' results in 29 who indicate that they recommend aspirin.
Determine the 99% confidence interval for the proportion of doctors who recommend aspirin.
Find the nearest answer.
[11.40%, 28.60%]
[9.68%, 30.32%]
[2.96%, 17.04%]
[10.62%, 29.38%]
[2.26%, 17.74%]
| a. | [78032, 81968] | |
| b. | [76130, 83870] | |
| c. | [77540, 82460] | |
| d. | [76904, 83096] | |
| e. | [77060, 82940] |
Solution
a)
CI = x ± Z a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=80000
Standard deviation( sd )=12000
Sample Size(n)=100
Confidence Interval = [ 80000 ± Z a/2 ( 12000/ Sqrt ( 100) ) ]
= [ 80000 - 1.64 * (1200) , 80000 + 1.64 * (1200) ]
= [ 78032,81968 ]
b)
Compute Sample Size
n = (Z a/2 * S.D / ME ) ^2
Z/2 at 0.01% LOS is = 2.58 ( From Standard Normal Table )
Standard Deviation ( S.D) = 0.25
ME =0.06
n = ( 2.58*0.25/0.06) ^2
= (0.645/0.06 ) ^2
= 115.5625 ~ 116
c)
Confidence Interval For Proportion
CI = p ± Z a/2 Sqrt(p*(1-p)/n)))
x = Mean
n = Sample Size
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=29
Sample Size(n)=145
Sample proportion = x/n =0.2
Confidence Interval = [ 0.2 ±Z a/2 ( Sqrt ( 0.2*0.8) /145)]
= [ 0.2 - 2.58* Sqrt(0.001) , 0.2 + 2.58* Sqrt(0.001) ]
= [ 0.114,0.286] ~ [ 11.40% to 28.60% ]

