# Statistics Essay Example

• Category:
Other
• Document type:
Math Problem
• Level:
• Page:
1
• Words:
648

7Statistics

Statistics

Question 1

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

2. Median, is the middle number = 39

3. Standard deviation, SN SN = √{∑f(x-x)2}/n-1

SN = √{44/4} = 3.32

1. 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) =

1. z = (x — µ)/σ, with µ = 90 and σ = 10

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

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

1. 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

1. P(X>85) for a sample of 25 cars; Z (-0.05) = 0.5199 ≈ 0.52

Question 4

1. A 90 % confidence interval for the true proportion of students with “Statisticitis”

Computing the test statistic 1. 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 t-test 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
1. indices for 2008-2009 price increases

1. 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

1. 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

1. Both Paasche and Laspeyeres indices are approximately equal.

Reference

Spiegel, M. R. (1992). Theory and Problems of Probability and Statistics. New York: McGraw-Hill, pp. 111-112.