Statistics Essay Example
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 Document type:Math Problem
 Level:Undergraduate
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 Words:648
7Statistics
Statistics
Question 1

Mean = (34+39+41+35+41)/5 = x = 38

Median, is the middle number = 39

Standard deviation, S_{N}
S_{N} = √{∑f(xx)^{2}}/n1
S_{N} = √{44/4} = 3.32

Mean x = ∑fx/n = 870/30 = x = 29
1.5 Median = L + [(n/2 – CF)/f]i = 20 + [(30/2 – 4)/30]10 = 23.67 ≈ 25
1.6 The first quartile of the sample Q1 = Q1 is the value that has 25% of the data set below it, and 75% above it.Q1 = (n + 1)1/4 = 7.75 class which is 20
Question 2
(a) 10% work in the finishing department; 20% are absent excessively; and 7% work in the finishing department and are absent excessively.
P(A) = 0.2; P (F) = 0.1 and P(PnF) = 0.07
2.1 P(AnF) = 0.07
2.2 P(AuF) = P(A) + P (F) — P(PnF)= 0.2 + 0.1 – 0.07= 0.23
2.3 P(A/F) = [P(A)P(F/A)]/P(A) = 0.02/0.2 = 0.1
P(F/A)/P(F)
P(A.F) = P(A)P(F/A) = P(F)P(A/F)
2.4 P(A/F’) = [P(A.F’)]/P(A) = 0.18/0.2 = 0.9
(b) Dr Smith sees 40%, Dr Tran sees 30% and Dr Jackson sees 30%. Dr Smith requests blood tests on 5% of her patients, Dr Tran 8% and Dr Jackson 6%.
2.5 Probability of a blood test, P(B) = 0.4*0.05 + 0.3*0.08 + 0.3*0.06 = 0.062
2.6 P(S/B) =
Question 3
P(X < 1) = P(X = 0) + P(X = 1) = 0.25 + 0.50 = 0.75
P (X<3) = P(X = 0) + P(X = 1) + P(X = 2) =

z = (x — µ)/σ, with µ = 90 and σ = 10

P(82<x<108), x1 = 0.8 and x2 = 1.8

Hence P(82<x<108) = Z whose value is 2.6 = 0.9974

P(Z whose value is 90%); z = 1.283
Hence 1.283*10 = x – 90
X = 102.83Km/h maximum speed without receiving a speeding ticket

P(X>85) for a sample of 25 cars; Z (0.05) = 0.5199 ≈ 0.52
Question 4

A 90 % confidence interval for the true proportion of students with “Statisticitis”
Computing the test statistic

σ = 6, to estimate a µ=<2 with a confidence level of 98%. What is the sample size?
Z = (x — µ)/( σ/√n), but at 98% confidence interval, the Z critical value is 2.067
Thus 2.067 = (x — 2)/(6/)
Question 5
Investment A µ = 8.8, µ = 3.12 and investment B µ = 7.48, µ = 4.05
At 5% significance level, observed Z statistic is 1.96
z = (x — µ)/σ,
hence for A, 1.96 = (x – 8.8)/3.12, therefore x = 14.92
and B, 1.96 = (x – 7.48)/4.05, and x = 15.42
thus, the average rate of return for investment A is not greater than the return for investment B.
Question 6
(a) From a simple excel regression analysis, the intercepts are 3434.79 and 2811.34 for Frontage, Hence Y = 2811.34×1 +3434.79
(b) At 1% significance the relationship between frontage and annual rent had a significant positive slope of 0.897
(c) Using Y = 2811.34×1 +3434.79, the annual rent of a mall with a frontage of 6.0 meters is 16868.04 dollars. The estimate is at 99% confidence interval as a single tail ttest is used.
Question 7
Mean and standard deviation
Salt and vinegar 

Cheese and onion 

Mean = 60 and standard deviation √(868/4) = 14.73
Z= (x — µ)/σ 

Salt and vinegar 

cheese and onion 
From the above observed Z values and at 1% significance level with critical Z value of +/2.576, the researcher can conclude that all the flavors are equally popular.
Question 8
Round Leg ham 
Chicken roll 


indices for 20082009 price increases

Paasche ‘s price index

Paasche Price Index =
= (5.5*560+19.98*750+11.98*650)/( 4.95*560+18.95*750+11.5*650) = 1.057

Laspeyre’s price index
Laspeyres Price Index =
= (5.5*550+19.98*650+11.98*500)/( 4.95*550+18.95*650+11.5*500) = 1.058

Both Paasche and Laspeyeres indices are approximately equal.
Reference
Spiegel, M. R. (1992). Theory and Problems of Probability and Statistics. New York: McGrawHill, pp. 111112.