In one study, out of $80$ respondents, $23$ were against going to a concert. Determine if a person is against decisions to go to the concert if the confidence interval is 95%.
Help me please. Thanks for your attention and your help.
My attemp is: from $n = 80$ respondents, against the sales of the government have been declared $m = 23$, for this the possibility of a person accidentally of the President will be against the decisions of the government is $\widehat p=\frac{m}{n}=\frac{23}{80}.$ Since, $n\widehat p=23\geq 10 $ and $n(1-\widehat p)=57\geq 10$ qe find the confidence interval we get the formula:
$$(\widehat p+z_{\frac{\alpha}{2}}\sqrt\frac{\widehat p(1-\widehat p)}{n},\widehat p+z_{\frac{\alpha}{2}}\sqrt\frac{\widehat p(1-\widehat p)}{n})$$
Then $1-\alpha=0.95\rightarrow \alpha =0.25$ and $1-\frac{\alpha}{2}=0.975.$
The probability $0.975$ corresponds to the $z$-value $1.96$, therefore $z_{\frac{\alpha}{2}}=z_{0.025}=1.96$.
I didnt know it is correct.
Unclear and confused terminology aside, the Wald-type confidence interval for a binomial proportion is given by $$\left( \hat p - z_{\alpha/2}^* \sqrt{\frac{\hat p(1-\hat p)}{n}}, \hat p + z_{\alpha/2}^* \sqrt{\frac{\hat p(1-\hat p)}{n}}\right),$$ where $\hat p = x/n$ is the observed proportion in the sample, $n$ is the sample size, $x$ is the observed number of events of interest, and $z_{\alpha/2}^*$ is the upper $\alpha/2$ quantile of the standard normal distribution, i.e. it is the number such that $$\Pr[Z > z_{\alpha/2}^*] = \alpha/2$$ where $Z \sim \operatorname{Normal}(0,1)$. For a confidence level of $95\%$, this corresponds to $\alpha = 0.05$, hence $$z_{0.025}^* \approx 1.96.$$ The observed proportion, or point estimate, is $$\hat p = \frac{23}{80} = 0.2875.$$ We compute the standard error $$SE = \sqrt{\frac{\hat p (1-\hat p)}{n}} = \sqrt{\frac{(0.2875)(1 - 0.2875)}{80}} \approx 0.0506018.$$ The margin of error is therefore $$ME = z_{\alpha/2}^* SE \approx (1.96)(0.0506018) \approx 0.0991778.$$ The lower confidence limit is $$LCL = \hat p - ME \approx 0.2875 - 0.0991778 \approx 0.188322,$$ and the upper confidence limit is $$UCL = \hat p + ME \approx 0.386678.$$ Consequently, our Wald-type interval estimate for the true proportion of respondents who are against going to a concert is $$(0.188322, 0.386678)$$ for a $95\%$ confidence level. This is not the only possible estimate; the Clopper-Pearson or Wilson score intervals are also computable and give slightly different estimates.