Question

In 2011, when the Gallup organization polled investors, 31% rated gold the best long-term investment. But in April of 2013 Gallup surveyed a random sample of U.S. adults. Respondents were asked to select the best long-term investment from a list of possibilities. Only 238 of the 950 respondents chose gold as the best long-term investment. By contrast, only 92 chose bonds.

a. Compute the standard error for each sample proportion. Compute and describe a 95% confidence interval in the context of the question.

b. Do you think opinions about the value of gold as a long-term investment have really changed from the old 31% favorability rate, or do you think this is just sample variability? Explain.

c. Suppose we want to increase the margin of error to 5%, what is the necessary sample size?

d. Based on the sample size obtained in part c, suppose 112 respondents chose gold as the best long-term investment. Compute the standard error for choosing gold as the best long-term investment. Compute and describe a 95% confidence interval in the context of the question.

e. Based on the results of part d, do you think opinions about the value of gold as a long-term investment have really changed from the old 31% favorability rate, or do you think this is just sample variability? Explain.

Answer 1

Solution

Part (a)

Let p_g and p_b be respectively the population proportion of investors who think gold and bond as the best long-term investment.

Let p_ghat and p_bhat be the corresponding sample proportions.

Then

p_ghat = 238/950

= 0.2505 Answer 1

p_bhat = 92/950

= 0.0968 Answer 2

Standard error for p_ghat = √{p_ghat(1 - p_ghat)/n}

= √(0.2505 x 0.7495/950)

= 0.0141 Answer 3

Standard error for p_bhat = √{p_bhat(1 – p_bhat)/n}

= √(0.0968 x 0.9032/950)

= 0.0096 Answer 4

95% confidence interval for population proportion, p_g is: [0.2230, 0.2781] Answer 5

95% confidence interval for population proportion, p_b is: [0.0781, 0.1156] Answer 6

Back-up Theory

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.

Calculations

For p_g

n	950
X	238
p' = phat	0.250526
F = p'(1-p')/n	0.000198
sqrtF	0.014059
α	0.05
1 - (α/2)	0.975
Zα/2	1.959964
MoE	0.027554
LB	0.222972
UB	0.278081

For p_b

n	950
X	92
p' = phat	0.096842
F = p'(1-p')/n	9.21E-05
sqrtF	0.009595
α	0.05
1 - (α/2)	0.975
Zα/2	1.959964
MoE	0.018806
LB	0.078036
UB	0.115648

Part (b)

This question is addressed with test of significance as described below:

Claim :

Opinions about the value of gold as a long-term investment have not really changed from the old 31%.

Hypotheses:

Null H₀ : p = p₀ = 0.31[claim] Vs Alternative H_A : p ≠ 0.31

Test Statistic:

Z = (p_hat - p₀)/√{p₀(1 - p₀)/n}

Where

p_hat = sample proportion and

n = sample size.

Calculations:

p0	0.31
n	950
x	238
phat	0.250526316
Zcal	-3.9635
α	0.05
Zcrit	1.9600
p-value	0.0001

Distribution, Significance Level, α Critical Value and p-value:

Under H₀, distribution of Z can be approximated by Standard Normal Distribution, provided

np₀ and np₀(1 - p₀) are both greater than 10.

So, given a level of significance of α%, Critical Value = upper (α/2)% of N(0, 1), and

p-value = P(Z > | Zcal |)

assuming α = 0.05

Using Excel Function: Statistical NORMINV and NORMDIST these are found as shown in the above table.

Decision:

Since | Zcal | > Zcrit, or equivalently, since p-value < α, H₀ is rejected.

Conclusion :

There is not enough evidence to suggest that the claim is valid and hence we conclude that

Opinions about the value of gold as a long-term investment have really changed from the old 31%.

Answer 7

Part (c)

Vide (2) under Part (a),

MoE = Z_α/2[√{p_hat (1 – p_hat)/n}]

We want this to be 0.05.

i.e., 1.96√(0.2505 x 0.7495/n) = 0.05

Or, n = 288.50

Thus, to increase the margin of error to 5%, the necessary sample size is 289 Answer 8

Part (d)

p_ghat = 112/289

= 0.3875 Answer 9

Standard error for p_ghat = √{p_ghat(1 - p_ghat)/n}

= √(0.3875 x 0.6125/289)

= 0.0287 Answer 10

95% confidence interval for population proportion, p_g is: [0.3314, 0.4437] Answer 11

Refer Part (a) for Back-up Theory

Calculations

n	289
X	112
p' = phat	0.387543
F = p'(1-p')/n	0.000821
sqrtF	0.028658
α	0.05
1 - (α/2)	0.975
Zα/2	1.959964
MoE	0.056169
LB	0.331374
UB	0.443712

Part (e)

Opinions about the value of gold as a long-term investment have really changed from the old 31%Answer 12

Refer Part (b) for Back-up Theory

Calculations

p0	0.31
n	289
x	112
phat	0.387543253
Zcal	2.8503
α	0.05
Zcrit	1.9600
p-value	0.0044

DONE