business statistics - assignment 2 by haris awang

MBA 6073 Business Statistics: Assignment 2

By: A. HARIS AWANG

MBA2016041001

[email protected]

Submitted to:

Mr. Demudu Naganaidu

Senior Vice President Operations

Tel: 03-9080 5888

30th July, 2016

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MBA6073 Business Statistics: Assignment 2. A. HARIS AWANG. Asia Metropolitan University. 2016

MBA6073 Business Statistics:

Assignment 2 A. HARIS AWANG

1. Toby’s Trucking Company determined that the distance traveled per truck per year is

nomally distributed, with mean of 50.0 thousand miles and a standard deviation of 12.0 thousand

miles. (15 MARKS)

X ~ N(50.0, 122)

µ = 50.0, σ = 12.0,

a. What proportion of trucks can be expected to travel between 34.0 and 50.0 thousand miles in

the year? (5)

Formula:

Z

X, a = 34.0, b = 50.0 in thousand miles.

P (34.0 < X < 50.0) = P(

a< z <

b) = F(

b) – F(

a)

= F(0.12

0.500.50 ) - F(

0.12

0.500.34 )

= F(0.00) – F(-1.33)

From the z-table, F(0.00) = 0.5000, F(-1.33) = 0.0918

= 0.5000 – 0.0918

= 0.4082

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b. What percentage of trucks can be expected to travel either below 30.0 or above 60.0

thousand miles in the year? (5)

a = 30.0, b 60.0 in thousand miles.

P(x < 30.0) + P(x > 60.0) = P(z <

a) + P(z >

b)

= P(z <0.12

0.500.30 ) + P(z >

0.12

0.500.60 ) = P(z < -1.67) + P(z > 0.83)

= 0.0475 + 0.2033 (from z-table, use z = -0.83 since it is the tail end)

= 0.2508

= 25.08%

c. How many miles will be travelled by at least 80% of the trucks? (5)

“At least” means more than 80% of the trucks or 20% of the lower tail of the curve.

P(x > z) = 0.2 (from the z-table, the closest value is 0.2005 which gives z = -0.84)

x = zσ + µ = (-0.84)(12.0) + 50.0

= 39.92 x 1,000 miles

= 39,920 miles

2. It is reported that the mean download time for H&R Block website, www.hrblock.com, was

2.5 seconds. Suppose that the download time for the H&R Block website was normally

distributed, with a standard deviation of 0.5 second. If you select a random sample of 30

download times, (15 MARKS)

X ~ N(2.5, 0.52)

µ = 2.5, σ = 0.5 in seconds, n = 30

a. What is the probability that the sample mean is less than 2.75 seconds? (3)

Formula:

Z X

X

,

X =

n

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P )75.2( x = P(z <X

X

) = P( z < (2.75 – 2.5)/(0.5/√30) ) = P (z < 2.74)

From the z-table, at z = 2.74

= 0.9969

b. What is the probability that the sample mean is between 2.70 and 2.90 seconds? (3)

P )90.270.2( x = P( z < (2.90 – 2.5)/(0.5/√30) ) - P( z < (2.70 – 2.5)/(0.5/√30) )

= P(z < 0.4/0.0913) – P(z < 0.2/0.0913) = P(z < 4.38) – P(z < 2.19)

= 0.9999 – 0.9857

= 0.0142

c. The probability is 80% that the sample mean is between what two values symmetrically

distributed around the population mean? (4)

If the area symmetrically distributed around the population mean is 80%, then we are looking

at the tails which are 10% on each side. This corresponds to the value of z1 and z2 = 0.1000

and 0.9000 respectively.

P (z < z1) = 0.1000. z1 = -1.28 (from z-table)

P (z < z2) = 0.9000. z2 = +1.28

x 1 = z1(0.0913) + 2.5 = -1.28(0.0913) + 2.5 = 2.38

x 2 = z2(0.0913) + 2.5 = 1.28(0.0913) + 2.5 = 2.62

Where x 1 & x 2 are the two values.

d. The probability is 90% that the sample mean is less than what value? (5)

P (z < z1) = 0.9000. z1 = +1.28

x 1 = z1(0.0913) + 2.5 = 1.28(0.0913) + 2.5 = 2.62

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3. In a tea-bag-filling operation, an important characteristic of interest for this process is the

weight of the tea in the individual bags. The data in the file is an ordered array of the weight in

grams of a sample of 50 tea bags produced during an eight-hour shift. (15 MARKS)

a. Is there evidence that the mean amount of tea per bag is different from 5.5 grams (use

α=0.01)? (10)

It is a two-tailed test where,

H0 : µ = 5.5

H1 : µ ≠ 5.5

α = 0.01, n = 50, s = 0.10583 (from SPSS)

Critical value t49, 0.005 = ± 2.678

σ is unknown, so use t statistic.

tn-1

n

s

X 0 = (5.5014 – 5.5)/(0.10583/√50) = 0.0935

Decision: Since –tn-1, α/2 < 0.0935 < –tn-1, α/2, H0 cannot be rejected.

There is not enough evidence to reject H0 or to conclude that the mean amount of tea

bag is different from 5.5 grams.

Reject H0 Reject H

0

/2=.005

-t n-1,α/2

Do not reject H0

0

/2=.005

-2.678 2.678

t n-1,α/2

0.0935

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b. Construct a 99% confidence interval estimate of the population mean amount of tea per bag.

Interpret this interval. (5)

t

n

s

X 0

X LCL= t(n

s) +µ0

= -2.678(0.10583/√50) +5.5 = 5.46

X UCL= t(n

s) +µ0

= +2.678(0.10583/√50) +5.5 = 5.54

With 99% confidence interval, we can conclude that the population mean amount of tea

per bag is somewhere between 5.46 and 5.54 grams or 5.46 < µ < 5.54.

4. Circulation is the lifeblood of the publishing business. The larger the sales of a magazine, the

more it can charge advertisers. Recently, a circulation gap has appeared between the publishers

report of magazines newsstand sales and subsequent audits by the Audit Bureau of Circulations.

The data in the file represent the reported and audited newsstand (in thousands) in 2001 for the

following 10 magazines: (15 MARKS)

Magazine Reported (X) Audited (Y)

YM 621.0 299.6

CosmoGirl 359.7 207.7

Rosie 530.0 325.0

Playboy 492.1 336.3

Esquire 70.5 48.6

TeenPeople 567.0 400.3

More 125.5 91.2

Spin 50.6 39.1

Vogue 353.3 268.6

Elle 263.6 214.3

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a. State the Simple Linear Regression Model. (2)

Yi = β0 + β1Xi + εi

Where

Yi is dependent variable,

β0 is population Y intercept,

β1 is population slope coefficient,

Xi is independent variable,

εi is random error.

b. State the Estimated Simple Linear Regression Equation. (2)

ŷi = b0 + b1xi

Where

ŷi is estimated (or predicted) y value for observation i,

b0 is estimate of the regression intercept,

b1 is estimate of the regression slope,

xi is value of x for observation i.

c. Identify the estimated value of coefficient based SPSS output in Appendix 1. (4)

0.572

d. Interpret the meaning of the slope, b1 in this problem. (4)

The results suggested that 1,000 increase in reported sales was followed by 572 increase

in audited sales.

e. Predict the mean audited newsstand sales of 400,000. (3)

ŷi = b0 + b1xi = 26,724 + 0.572(400,000)

= 255,524

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5. (10 MARKS)

a. Summarize the forms for Null and Alternative Hypothesis about a Population Mean. (5)

b. Summarize the Hypothesis Test About the Difference between the Means of Two

Populations: Independent Samples. (5)

-END-

The equality part of the hyphothesis always appears in the null hypothesis.

A hypothesis test about the value of a population mean µ must take one of the

following three forms (where µ0 is the hypothesized value of the population mean).

H0: µ ≥ µ0

H1: µ < µ0

One-tailed

(lower-tail)

H0: µ ≤ µ0

H1: µ > µ0

One-tailed

(upper-tail)

H0: µ = µ0

H1: µ ≠ µ0

Two-tailed

The equality part of the hyphothesis always appears in the null hypothesis.

A hypothesis test about the value of the difference between two population means µ1

& µ2 must take one of the following three forms (where µ1 & µ2 are means from

independent samples).

H0: µ1 - µ2 ≥ 0

H1: µ1 - µ2 < 0

One-tailed

(lower-tail)

H0: µ1 - µ2 ≤ 0

H1: µ1 - µ2 > 0

One-tailed

(upper-tail)

H0: µ1 - µ2 = 0

H1: µ1 - µ2 ≠ 0

Two-tailed

business statistics - assignment 2 by haris awang

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