# Sampling Theory

Given a normally distributed population, a sampling distribution is summarized by Î¼X = 300 and ÏƒX = 28. What is the probability that a simple random sample will reveal a sample mean a. below 200? b. between 200 and 265? c. above 346?

Given â€“
Sample mean = 300
Standard error = 28

a) P(XÌ…<200)
= P(XÌ…-300/28 < 200-300/28)
= P(z â‰¤ -3.57)
= 0.000178

b) P (200 = P ( XÌ… < 265 ) â€“ P( XÌ… < 200 )
= P ( XÌ…-300/28 < 265-300 /28) â€“ P( XÌ… - 300 /28 < 200 -300/28)
= P (z<-1.25) â€“ P(z< -3.57)
= P(z>1.25) â€“ P(z> 3.57)
= [1 â€“ P(z<1.25)] â€“ [( 1 â€“ P(z<3.57) )]
= (1 - .8944) â€“ (1 -0.9998)
= 0. 1054

C) P(XÌ… > 346)
= 1- P(XÌ… â‰¤ 346)
= 1- P( (346 â€“ 300)/28)
= 1- P (1.643)
= 0.050191

A normally distributed population of 12 oz. beer cans has a mean of 105 calories and a standard deviation of 3 calories. Compute the mean of a sampling distribution for a simple random sample of 30 as well as the associated standard error.

Population mean = 105 oz
Population standard deviation = 3 oz
n = 30 oz
Mean of sampling distribution = E(XÌ… ) = Âµ = 105 oz
Standard error = Standard deviation / âˆšn
= 3/âˆš30
= 0.5477

The daily lunch expense of corporate executives is a normally distributed random variable with a mean of Â£9.9 and a standard deviation of Â£1. Describe the sampling distribution of the mean for a simple random sample of n = 36. What is the probability of encountering a sample mean between Â£10 and Â£11?

Probability of encountering a sample mean between Â£10 and Â£11 = P(10â‰¤xÌ…â‰¤11)
= P(10 â€“ E(XÌ…)/Standard error â‰¤ xÌ… - E(XÌ…)/Standard error â‰¤ 11 - E(XÌ…)/Standard error )
= Standard error = Standard deviation / âˆšn
= 1/âˆš36
= 1/6
= 0.16667
P (10 â€“ 9.9/ 0.16667 â‰¤ xÌ… - 9.9/ 0.16667 â‰¤ 11- 9.9/ 0.16667)
= P (0.599 â‰¤ z â‰¤ 6.59)
= P (zâ‰¤6.59) â€“ P (zâ‰¤0.599)
= 1- 0.274586 (since for any Z greater than 4 we assume Probability to be equal to 1)
= 0.725414 or 72.54%

An investigator knows that the population of light bulb lifetimes is normally distributed and that the standard deviation is 36 hours. A simple random sample of 49 bulbs discloses a sample mean lifetime of 510 hours. Determine a 95% confidence interval for the mean lifetime of all such bulbs.

Population standard deviation = 36
n = 49
XÌ… = 510
Confidence level = 95% = 1 - Î±
We know
Âµ = XÌ… Â± ZÎ±/2 * Standard Error of XÌ…
Z_(Î±/2)=1.96 (from normal distribution table)
Standard Errors of XÌ… = Population standard deviation/âˆšn
= 36/âˆš49
= 5.1428
It implies Âµ = 510 Â± 1.96 (5.1428)
= 499.92 â‰¤ Âµ â‰¤ 520.08

A banker who believes the relevant statistical population to be normally distributed wants to estimate the average monthly mortgage payment in a city. A simple random sample of 28 homeowners produces a mean of Â£426.73 with a standard deviation of Â£29.75. Determine a 99.8% confidence interval for the citywide average monthly payment.

Sample standard deviation = S = 29.75
n = 28
XÌ… = 426.75
Confidence level = 99.8% = 1 â€“ Î±
We know
Âµ = XÌ… Â± ZÎ±/2 * Standard Error of XÌ…
ZÎ±/2=0.5839(from normal distribution table )
Standard Error of XÌ… = Sample standard deviation/âˆšn
= 29.75/âˆš28
= 5.6627
It implies Âµ = 426.73 Â± 0.5839(5.6627)
= 410.25 â‰¤ Âµ â‰¤ 444.10

Passengers abandoning a sinking liner have a mean weight of 77kg, with a standard variation of 8kg. 40 passengers clamber onto a lifeboat, which will capsize if the total weight of the passengers exceeds 3,200kg. What is the probability that the lifeboat will capsize?

Population mean = Âµ = 77 kg = E(xÌ…)
Population Standard deviation = 8 kg
Standard error = Population Standard deviation/ âˆšn
Total passengers = 40
Average weight of 1 passenger for lifeboat to capsize with 40 persons = 3200/40 = 80
Standard Error of weights of Passenger
= 8/âˆš40
= 1.265
n = 40
Probability of boat capsizing = P (XÌ… â‰¥ 3200)
= P (XÌ… - Âµ /Standard error of XÌ… â‰¥ 3200 - Âµ/Standard error of XÌ…)
= P (Z â‰¥ 80 â€“ 77/1.265)
= P(z â‰¥ 2.37)
= 1-P( z â‰¤2.37)
= 1 - .99111
= 0.00889

A random sample of 65 weekly rents for one-person flats provides the following information: sample mean X = Â£345, sample deviation s = Â£32. Find a 98% confidence interval for the true population mean rent. Interpret your result carefully

Sample standard deviation = s = 32
n = 65
XÌ… = 345
Confidence level = 98% = 1 - Î±
We know
Âµ = XÌ… Â± Z_(Î±/2) * Standard Error of XÌ…
ZÎ±/2=0.2885( from normal distribution table )
Standard Error of XÌ… = Sample standard deviation/âˆšn
=32 /âˆš65
= 3.96
It implies Âµ = 345 Â± 0.885 ( 3.96 )
= 335.766 â‰¤ Âµ â‰¤ 354.3323
Interpretation: This means that in 98 out of 100 one-person flats sampled, the mean weekly rent will lie in between Â£336 and Â£355

A sample of 100 supermarket shoppers spend on average Â£56 with a standard deviation of s = Â£10. Find a 95% confidence interval for the true mean of shoppersâ€™ spending.

Sample standard deviation = S = 10
n = 100
XÌ… = 56
Confidence level = 95% = 1 - Î±
We know
Âµ = XÌ… Â± ZÎ±/2 * Standard Error of XÌ…
ZÎ±/2 = 1.96( from normal distribution table ) Standard Error of XÌ… = Sample standard deviation/âˆšn
=10 /âˆš100
= 1
It implies Âµ = 56Â± 1.96 (1)
= 54.04 â‰¤ Âµ â‰¤ 57.96

Suppose two economists estimate Î¼ (the average expenditure of British families on food) with two unbiased (and statistically independent) estimates U and V. The second economist is less careful than the first: the standard deviation of V is 3 times as large as the standard deviation of U. When asked how to combine U and V to get a publishable overall estimate, three proposals are made: â€¢ W1 = Â½ U + Â½ V (simple average) â€¢ W2 = Â¾ U + Â¼ V (weighted average) â€¢ W3 = 1 U + 0 V (drop the less accurate estimate) (a) Which are unbiased? (b) Which is the best estimator? The worst? (c) Is it possible to find an even better estimator?

W1 = Â½ U + Â½ V (simple average)
Taking expectation both sides
It implies, E(W1) = Â½ E(U) + Â½ E(V)
= Â½ Âµ + Â½ Âµ = Âµ
And E(W1) also equals Âµ
LHS = RHS
Therefore W1 is unbiased
W2 = Â¾ U + Â¼ V (weighted average)
Taking Expectation both sides
RHS
E(3/4U+1/4V)
=3/4*E(U)+1/4*E(V)
=3/4Âµ+1/4*Âµ
=Âµ
LHS
E(W2) =Âµ
HENCE LHS = RHS
SO W2 IS UNBIASED
W3 = 1 U + 0 V (drop the less accurate estimate)
Taking Expectation both sides
RHS
E(1U+0V)
=1*E(U)+0*E(V)
=Âµ
LHS
E(W3)
=Âµ
LHS=RHS
HENCE W3 IS UNBIASED

B) W1 = Â½ U + Â½ V (simple average)
TAKING VARIANCE BOTH SIDES

RHS
V(1/2U+1/2V)
=1/2*1/2V(U)+1/2*1/2V(V)
=1/4V(U)+1/4(9V(U))
=5/2V(U)

W2 = Â¾ U + Â¼ V (weighted average)

TAKING VARIANCE BOTH SIDES
RHS
V( Â¾ U + Â¼ V)
=3/42V(U)+1/42V(V)
=9/16V(U)+1/16(9V(U))
=9/8V(U)

W3 = 1 U + 0 V (drop the less accurate estimate
RHS
V(1U+0V)
=1V(U)

It implies W3 is best estimator as it has less variance Also W1 is worst estimator as it has most variance (By Minimum Variance Unbiased Estimator)

(Estimating a population proportion) A random sample of 100 workers in an economy reveals that 20% are unemployed. Obtain an unbiased point estimate of the true population unemployment level. Hence obtain a 99% confidence interval for the true population unemployment level.

n = 100
pË† = 20% = 0.2
1-Î± = 99%
We know
P = pË† Â± ZÎ±/2 SD (pË†)
Where SD (pË†) is

By putting the values it implies
SD (pË†) = 0.04
ZÎ±/2 = 2.575
Putting the value in formula 1 we get
P = 0.2 Â± 2.575(0.04)
0.097 â‰¤ P â‰¤ 0.303

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