The confidence interval for population mean is given by :-
[tex]\hat{p}\pm E[/tex], where [tex]\hat{p}[/tex] is sample proportion and E is the margin of error .
[tex]E=z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex]
Given : Significance level : [tex]\alpha:1-0.95=0.05[/tex]
Sample size : n= 64
Critical value : [tex]z_{\alpha/2}=1.96[/tex]
Sample proportion: [tex]\hat{p}=\dfrac{22}{64}\approx0.344[/tex]
[tex]E=(1.96)\sqrt{\dfrac{0.344(1-0.344)}{64}}\approx0.1164[/tex]
Hence, the margin of error = 0.1164
Now, the 95% theory-based confidence interval for the population proportion will be :
[tex]0.344\pm0.1164\\\\=(0.344-0.1164,\ 0.344+0.1164)=(0.2276,\ 0.4604)[/tex]
Hence, the 99% confidence interval is [tex](0.2276,\ 0.4604)[/tex]
