Question

In: Statistics and Probability

When the gallup organization polled investors, 34% rated gold the best long-term investment.... Question: In 2011,...

When the gallup organization polled investors, 34% rated gold the best long-term investment.... Question: In 2011, when the Gallup organization polled investors, 34% rated gold the best long-term investm... In 2011, when the Gallup organization polled investors, 34% 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 241 of the 1005 respondents chose gold as the best long-term investment. By contrast, only 91 chose bonds.

Could this be explained using Excel ? each explanation in each spread sheet? or in one.

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 34% favorability rate, or do you think this is just sample variability? Explain.

c. Suppose we want to increase the margin of error to 3%, what is the necessary sample size?

d. Based on the sample size obtained in part c, suppose 120 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 34% favorability rate, or do you think this is just sample variability? Explain.

Solutions

Expert Solution

TRADITIONAL METHOD
given that,
possibile chances (x)=241
sample size(n)=1005
success rate ( p )= x/n = 0.24
I.
sample proportion = 0.24
standard error = Sqrt ( (0.24*0.76) /1005) )
= 0.013

II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
margin of error = 1.96 * 0.013
= 0.026
III.
CI = [ p ± margin of error ]
confidence interval = [0.24 ± 0.026]
= [ 0.213 , 0.266]
-----------------------------------------------------------------------------------------------
DIRECT METHOD

given that,
possibile chances (x)=241
sample size(n)=1005
success rate ( p )= x/n = 0.24
CI = confidence interval
confidence interval = [ 0.24 ± 1.96 * Sqrt ( (0.24*0.76) /1005) ) ]
= [0.24 - 1.96 * Sqrt ( (0.24*0.76) /1005) , 0.24 + 1.96 * Sqrt ( (0.24*0.76) /1005) ]
= [0.213 , 0.266]
-----------------------------------------------------------------------------------------------
interpretations:
1. We are 95% sure that the interval [ 0.213 , 0.266] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population proportion

[ANSWERS]
a. standard error = 0.013
b. [0.213 , 0.266]
c. long term investment value is changed from old 34% since it does n;t lies in the
interval achived


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