Question

In a past​ election, the voter turnout was 69%

In a​ survey, 930 subjects were asked if they voted in the election.

a. Find the mean and standard deviation for the numbers of voters in groups of 930

b. In the survey of 930 ​people 621said that they voted in the election. Is this result consistent with the​ turnout, or is this result unlikely to occur with a turnout of 69​%?

Why or why​ not?

c. Based on these​ results, does it appear that accurate voting results can be obtained by asking voters how they​ acted?

Answer 1

a) n = number of persons asked if they voted in the election. = 930.

P : Population proportion of persons they voted in the election = 0.69

X : Number of persons are voted in the election.

The probability distribution of random variable X is binomial with parameter n = 930 and P = 0.69

X ~ Bin ( n= 930, P = 0.69)

E(X) = mean = nP = 930 *0.69 = 641.7

SD(X) =sqrt(n*P*(1-P)) = sqrt( 930 * 0.69 *0.31) = 14.1041

b) Since P =population proportion of persons voted in the election = 0.69

p : Sample proportion of persons voted in the election.

X : Number of person voted in the election = 621

p= X/ n = 621 / 930 = 0.6677

We have to test the hypothesis

Whether or not sample proportion is differ from population proportion?

i.e. Null Hypothesis- Ho : P = 0.69

against

Alternative Hypothesis- Ha: P # 0.69 (two-tailed test)

We used one sample proportion test.

The value of test statistic is

The value of test statistic Z = -1.4671

Consider

Alpha : level of significance = 0.05

Since value of test statistic is -1.4671 and test is two-tailed p-value is obtained by

from normal probability table

P (Z < -1.4671) = 0.0712

p-value = 0.1424

Since p-value > level of significance, we failed to reject Ho.

The result is consistent with the turnout.

c) From part(b) we accept the null hypothesis.

There is sufficient evidence support to claim that accurate result can be obtained by asking voters how they acted.