Question

In a certain hypothetical earthly country and epoch, among a simple random sample of 652 of the country’s population of adults who were fluent in just one human language, a point estimate of 52 percent said they had not learned any other human language because there was no need to – because their native tongue was very well used by most people in most of the world’s countries, and because everyone they interacted with also was fluent in that particular language.

At that time, an upstart digital medium reported that a majority of the country’s adults fluent only in the native language saw no need to learn any other language. The report quoted the 52 percent figure for evidence of this statement.

(i)

Perform a hypothesis test to uncover whether the data at hand truly support the digital medium’s claim. What do we conclude?

(ii)

Construct a 95% confidence interval for the proportion of the country’s adults who expressed a perceived lack of need to learn any other human language.

(iii)

If we seek a margin of error for the 95% confidence level to be 1.3 percent, how large should the sample size for a survey such as this reasonably be, using the point estimate of 0.52?

Among answer choices a. – d. that follow, CHOOSE THE BEST ANSWER.

a.

(i)

We conclude nothing: the requisite success-failure conditions fail to obtain in this case.

(ii)

None of the above intervals is correct.

(iii)

There is no correct size.

b.

(i)

We conclude the digital medium’s claim is sound – the evidence for a majority is strong.

(ii)

(0.515, 0.525)

(iii)

5,683

c.

(i)

We conclude ambiguity: both the medium’s claim and any counterclaim that might be made has statistical evidence, based on the results of the hypothesis test.

(ii)

(0.501, 0.520)

(iii)

5,627

d.

(i)

We conclude the digital medium’s claim is unsound – evidence for a majority is weak.

(ii)

(0.482, 0.558)

(iii)

5,674

e. None of the above answers sets is correct.

Answer 1

i) n = 652, p̄ = 0.52

Null and Alternative hypothesis:

Ho : p = 0.5 , H1 : p > 0.5

Test statistic:

z = (p̄ -p)/√(p*(1-p)/n) = (0.52 - 0.5)/√(0.5 * 0.5/652) = 1.0214

p-value = 1- NORM.S.DIST(1.0214, 1) = 0.1535

Decision:

p-value > α, Do not reject the null hypothesis

ii) 95% Confidence interval :

At α = 0.05, two tailed critical value, z_c = NORM.S.INV(0.05/2) = 1.960

Lower Bound = p̄ - z_c*√( p̄ *(1- p̄ )/n) = 0.52 - 1.96 *√(0.52*0.48/652) = 0.482

Upper Bound = p̄ + z_c*√( p̄ *(1- p̄ )/n) = 0.52 + 1.96 *√(0.52*0.48/652) = 0.558

(0.482, 0.558)

iii) Proportion, p = 0.52

Margin of error, E = 0.013

Confidence Level, CL = 0.95

Significance level, α = 1 - CL = 0.05

Critical value, z = NORM.S.INV(0.05/2) = 1.9600

Sample size, n = (z² * p * (1-p)) / E² = (1.96² * 0.52 * 0.48)/ 0.013²

= 5673.5392 = 5674

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Answer D is correct