Question

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?

Confidence intervals
ANSWERED
asked 2021-07-30
A paper reported a \(\displaystyle{\left({1}-\alpha\right)}\) 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 \(\displaystyle\alpha\). If you want to test the hypothesis that the proportion of voters is greater than \(\displaystyle{65}\%\) at \(\displaystyle{1}\%\) significance, find \(\displaystyle{z}_{{{c}{a}{l}{c}}}\) value for this problem? Please report your answer to 2 decimal places.

Expert Answers (1)

2021-08-09

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,
\(\displaystyle\hat{{{p}}}={\frac{{{0.561}+{0.599}}}{{{2}}}}\)
\(\displaystyle={0.58}\)
Length of CI is expressed as difference between given proportion.
Margin of error is expressed as,
Margin of error(E) \(\displaystyle={\frac{{{1}}}{{{2}}}}\times\) length of CI
\(\displaystyle={\frac{{{1}}}{{{2}}}}\times{\left({0.599}-{0.561}\right)}\)
\(\displaystyle={\frac{{{1}}}{{{2}}}}\times{0.038}\)
\(\displaystyle={0.019}\)
Step 2
Hence the CI of the given proportion with margin of error is,
\(\displaystyle\hat{{{p}}}+{E}={0.58}+{0.019}={0.599}\)
\(\displaystyle\hat{{{p}}}-{E}={0.58}-{0.019}={0.561}\)
Hence, the proportions are (0.599,0.561)
The expression for standard deviation is,
\(\displaystyle{S}.{E}=\sqrt{{{\frac{{\hat{{{p}}}{\left({1}-\hat{{{p}}}\right)}}}{{{n}}}}}}\)
\(\displaystyle=\sqrt{{{\frac{{{0.58}\times{0.42}}}{{{2056}}}}}}\)
\(\displaystyle=\sqrt{{{\frac{{{0.2436}}}{{{2056}}}}}}\)
\(\displaystyle=\sqrt{{{0.00011848}}}\)
\(\displaystyle={0.01088}\)
Step 3
As know that,
Margin of error \(\displaystyle{\left({E}\right)}={z}_{{\frac{\alpha}{{2}}}}\times{S}.{E}\)
Hence, the value of z is,
\(\displaystyle{z}_{{\frac{\alpha}{{2}}}}={\frac{{{E}}}{{{S}.{E}}}}\)
\(\displaystyle={\frac{{{0.019}}}{{{0.01088}}}}\)
\(\displaystyle\approx{1.7463}\)
Thus
\(\displaystyle\frac{\alpha}{{2}}={P}{\left({Z}{<}-{1.7463}\right)}\)
\(\displaystyle\frac{\alpha}{{2}}={0.0274}\)
\(\displaystyle\alpha={0.0548}\)
Hence the confidence level is \(\displaystyle{\left({1}−\alpha\right)}={0.9452}\).
Step 4
Now, to find null hypothesis:
\(\displaystyle{H}_{{{0}}}:{p}={65}\)
\(\displaystyle{H}_{{{0}}}:{p}{>}{65}\)
Here, sample proportion is \(\displaystyle{0.58},{n}={2056}\) and claimed proportion is 0.65 at significance level \(\displaystyle\alpha={0.01}\).
Hence the standard deviation at 0.65 claimed proportion,
\(\displaystyle{S}.{E}=\sqrt{{{\frac{{\text{claimed proportion(1-claimed proportion)}}}{{{n}}}}}}\)
\(\displaystyle=\sqrt{{{\frac{{{0.65}\times{0.35}}}{{{2056}}}}}}\)
\(\displaystyle=\sqrt{{{\frac{{{0.2275}}}{{{2056}}}}}}\)
\(\displaystyle\approx{0.0105}\)
So the value of \(\displaystyle{z}_{{{c}{a}{l}{c}}}\) is:
\(\displaystyle{z}_{{{c}{a}{l}{c}}}={\frac{{\hat{{{p}}}-{0.65}}}{{{0.0105}}}}\)
\(\displaystyle={\frac{{{0.58}-{0.65}}}{{{0.0105}}}}\)
\(\displaystyle={\frac{{-{0.07}}}{{{0.0105}}}}\)
\(\displaystyle=-{6.6667}\)
Hence, the value of z is -6.67.

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