Question

Use the data below to calculate the paired-difference variable. Pair Observation Form Population 1 Population 2 21 24 26 Let the alternative hypothesis be that .> 2. Use the paired t-test to perform the required hypothesis test at the 5% significance level. Assume that in Milwaukee County, 42% of the homes available on the market are three bedroom homes. There are 34 3-bedroom homes currently on the market in West Allis. There are 61 homes listed in West Allis Calculate the p-hat value to find a 3 bedroom home in West Allis. Calculate the standard deviation of the p-hat. Create a 95% confidence interval for the proportion, p, of all 3-bedroom homes on the market in Milwaukee County What would we want the sample size to be if we wanted a margin of error to be 0.037 while using the West Allis data?

Answer 1

ANSWER:

Given that,

I. Paired t-test

a)

I. Paired t-test

Part (a)

Pair	Observation From		d
	Population 1	Population 2
1	21	23	-2
2	23	21	2
3	24	22	2
4	18	21	-3
5	28	29	-1
6	24	20	4
7	26	21	5

Ans 1

Part (b)

Paired t-test

Claim: µ₁ > µ₂ i.e., µ_d > 0

Hypotheses:

Null: H₀: µ_d = µ₀= 0 Vs Alternative: H_A: µ_d > 0 [claim]

Test Statistic:

t = (√n) (Dbar - µ₀)/s i.e., (√n) (dbar/s) = 0.866

where

d-bar and s are respectively, sample average and sample standard deviation based on n observations on d.

Calculations:

Summary of Excel calculations is given below:

n	7
dbar	1.0
s	3.055050463
√n	2.645751311
tcal	0.866025404
α	0.05
p-value	0.209876543
tcrit)	1.943180274

Distribution, Significance level (α), Critical Value and p-value:

Under H₀, t ~ t_{n - 1}. Hence, for level of significance α%, Critical Value = upper α% point of  t_{n - 1} and p-value = P(t_{n - 1} > tcal).

α = 5% [i.e., 0.05 (given)

Using Excel Function: Statistical TINV TDIST, the above are found to be as given in the above table.

Decision:

Since tcal < tcrit, or equivalently, since p-value > α, H₀ is accepted.

Conclusion:

There is not sufficient evidence to suggest that the claim is valid and hence we conclude that the two populations are not different. Ans 2

II .Milwaukee Problem:

Part (a)

p_hat = proportion of 3-bed homes in West Allis = 34/61 = 0.5574 Answer 2

Part (b)

Standard deviation of p_hat

= √{ p_hat(1 - p_hat)/n}

= √(0.5574 x 0.4426/61)

= 0.064 Ans 3

Part (c)

100(1 - α) % Confidence Interval for the population proportion, p is:

p_hat ± MoE, ………………...................................................…………………………………………………. (1)

where

MoE = Z_α/2[√{p_hat (1 – p_hat)/n}] ……….........................................……………….......……………………..(2)

with

Z_α/2 is the upper (α/2)% point of N(0, 1),

p_hat = sample proportion, and

n = sample size.

So, 95% Confidence Interval for the population proportion, p is [0.125, 0.433] Answer 4

Details

n	61
X	34
p' = phat	0.557377
F = p'(1-p')/n	0.004044
sqrtF	0.063596
α	0.05
1 - (α/2)	0.975
Zα/2	1.959964
MoE	0.124645
LB	0.432732
UB	0.682022

Part (d)

We want MoE = 0.037

i.e., vide (2),

Z_α/2[√{p_hat (1 – p_hat)/n}] = 0.037

Substituting the values,

n = (1.96/0.037)² x (0.5574 x 0.4426)

= 692.21

Thus, the required sample size = 693 Ans 5

DONE

[Going beyond,

Part (d) can easily be obtained by the following result:

If n₁ and n₂ are the required sample sizes to meet error levels of E₁ and E₂ respectively, then

(n₁/n₂) = (E2/E1)²

Or, n₂ = n₁/(E2/E1)²

In the given case, n₁= 61, E₁= 0.1246 and E₂ = 0.037

So, n₂ = 69