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Q1)In opposing the establishment of a new bank in a community an existing bank stated th
the proposed bank contained 75 percent of its accounts. An economic consulting firm r
households to determine the sample proportion, p_hat, which had accounts at the exis
225. (Round your answer(s) to 3 decimal places.)
Find the probability that p_hat is:
a. Less than 69 percent. =
b. More than 78 percent. =
Answers –
a)
Normal Distribution
Mean ( u ) =0.75
Standard Deviation ( sd )=Sqrt(P*Q/n) = 0.0289
Normal Distribution = Z= X- u / sd ~ N(0,1)
P(X < 0.69) = (0.69-0.75)/0.0289
= -0.06/0.0289= -2.0761
= P ( Z <-2.0761) From Standard Normal Table
= 0.0189
b)
P(X > 0.78) = (0.78-0.75)/0.0289
= 0.03/0.0289 = 1.0381
= P ( Z >1.038) From Standard Normal Table
= 0.1496
Q2) Determine the p-value associated with each of the following values of the standard
your answer(s) to 3 decimal places.)
a. Two-tail test, with z = 1.5
p-value =
b. One-tail test, with z = 1.05
p-value =
c. One-tail test, with z = -2.4
p-value =
Answers –
a) P-Value : Two Tailed ( double the one tail ) - Ha : ( P != 1.5 ) = 0.134
b) P-Value : Right Tail - Ha : ( P > 1.05 ) = 0.147
C) P-Value : Left Tail - Ha : ( P < -2.4 ) = 0.008
,Zα/2 at 0.02% LOS is = 2.33 ( From Standard Normal Table )
Standard Deviation ( S.D) = 15
ME =4
n = ( 2.33*15/4) ^2
= (34.95/4 ) ^2
= 76.3439 ~ 77
Q5) An airline, in response to passenger concern, conducts realistic tests to determine
Boeing 737. Average evacuation time in the past has been 100 seconds with a standard
FFA requires tests that show a planeload of passengers can be evacuated in 98 second
sample of 30 tests, what is the probability that the average evacuation time will be 98
answer(s) to 3 decimal places.)
P(x-bar <=98) =
Answers –
Normal Distribution
Mean ( u ) =100
Standard Deviation ( sd )=15
Number ( n ) = 30
Normal Distribution = Z= X- u / (sd/Sqrt(n) ~ N(0,1)"
"P(X > 98) = (98-100)/15/ Sqrt ( 30 )
= -2/2.739= -0.7303
= P ( Z >-0.7303) From Standard Normal Table
= 0.7674
P(X<=98) = 1 - 0.7674 = 0.2326
Q6) Women make up 58% of the U. S. civilian workforce of 124 million. The U. S. Depar
selects 100 workers for a conference on national health care. What is the probability
workers are female? (Round your answer(s) to 4 decimal places.)
P(x > 59) =
Answers –
Normal Distribution to Binomial Approximation
Mean ( u ) = 100 * 0.58 = 58
Standard Deviation ( sd )= Sqrt( 100 * 0.58 * 0.42) = 4.936
Normal Distribution = Z= X- u / sd ~ N(0,1)
P(X > 59) = (59-58)/4.936
= 1/4.936 = 0.2026
= P ( Z >0.203) From Standard Normal Table
= 0.4197
Q7)
, = P ( Z <-0.875) From Standard Normal Table
= 0.19079
P(395 < X < 500) = 0.19079-0.00023 = 0.1906
Q8)
The dimensions of a two by four board are not actually two inches by four inches. Rath
supposed to be 1.50 inches by 3.50 inches. A planer finishes the boards to the proper
control, periodic checks are made to see that the planer is set to the correct dimensio
properly, the average width is μ= 3.5 inches. If the average width is not μ not = 3.5 i
adjusted. Thus the null hypothesis is μ = 3.5 inches and the alternate hypothesis is
purposes, a random sample of 43 boards is taken twice daily. The standard deviation o
inches. To minimize the unnecessary adjustment of the planer, the risk of a Type I er
the lower and upper values for the confidence interval around x. (Round your answer(s
Lower value =
Upper value =
Answers –
Confidence Interval
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)=3.5
Standard deviation( sd )=0.27
Sample Size(n)=43
Confidence Interval = [ 3.5 ± Z a/2 ( 0.27/ Sqrt ( 43) ) ]
= [ 3.5 - 1.96 * (0.0412) , 3.5 + 1.96 * (0.0412) ]
= [ 3.4193,3.5807 ]
Lower value = 3.420
Upper value = 3.581
Q9)
There are 300 students enrolled in Business Statistics. Historically, exam scores are
standard deviation of 33.15. Your instructor randomly selected a sample of 50 examin
Determine
Answers –
Mean(x)=71.9
Standard deviation( sd )=33.15
Sample Size(n)=300
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