A paper reported a (1-\alpha) confidence interval for the proportion of voters is (0.561,0.599) based on a sample of 2,056 people. However, the paper omitted the value of \alpha. If you want to test the hypothesis that the proportion of voters is greater than 65\% at 1\% significance, find z_{calc} value for this problem?

UkusakazaL

UkusakazaL

Answered question

2021-07-30

A paper reported a (1α) confidence interval for the proportion of voters is (0.561,0.599) based on a sample of 2,056 people. However, the paper omitted the value of α. If you want to test the hypothesis that the proportion of voters is greater than 65% at 1% significance, find zcalc value for this problem? Please report your answer to 2 decimal places.

Answer & Explanation

ka1leE

ka1leE

Beginner2021-08-09Added 1 answers

Given data,
Total number of people is 2056.
The proportion of voters is (0.561, 0.599)
Step 1
This is a symmetrical CI.
Hence the sample proportion is expressed as,
p^=0.561+0.5992
=0.58
Length of CI is expressed as difference between given proportion.
Margin of error is expressed as,
Margin of error(E) =12× length of CI
=12×(0.5990.561)
=12×0.038
=0.019
Step 2
Hence the CI of the given proportion with margin of error is,
p^+E=0.58+0.019=0.599
p^E=0.580.019=0.561
Hence, the proportions are (0.599,0.561)
The expression for standard deviation is,
S.E=p^(1p^)n
=0.58×0.422056
=0.24362056
=0.00011848
=0.01088
Step 3
As know that,
Margin of error (E)=zα2×S.E
Hence, the value of z is,
zα2=ES.E
=0.0190.01088
1.7463
Thus
α2=P(Z<1.7463)
α2=0.0274
α=0.0548
Hence the confidence level is (1α)=0.9452.
Step 4
Now, to find null hypothesis:
H0:p=65
H0:p>65
Here, sample proportion is 0.58,n=2056 and claimed proportion is 0.65 at significance level α=0.01.
Hence the standard deviation at 0.65 claimed proportion,
S.E=claimed proportion(1-claimed proportion)n
=0.65×0.352056
=0.22752056
0.0105
So the value of zcalc is:
zcalc=p^0.650.0105
=0.580.650.0105
=0.070.0105
=6.6667
Hence, the value of z is -6.67.

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