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
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.

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.