In a survey of 500 people selected at random, 345 people are asked if they love to travel alone.

(a) What is the estimated proportion of the population that love to travel alone? $$\hat{p}=\dfrac{X}{n}=\dfrac{345}{500}=0.69$$ (b) Calculate the $95\%$ confidence interval for this proportion. $$\hat{p}\pm z_{\frac{\alpha}{2}}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$$ $$0.68\pm 1.96\sqrt{\dfrac{0.68(1-0.68)}{500}}$$ $$0.68\pm 0.0409$$ The confidence interval is (0.6391, 0.7209).

The following month another group of 500 random people was asked the same question and 300 said they prefer to travel alone. (c) what is the estimated change in the population that prefer to travel alone. $$\dfrac{345-300}{500}=0.09$$ $\textbf{please is this approach correct ?}$

(d) calculate directly a $99\%$ confidence interval for this change using the asymptotic normal distribution.

$\textbf{please I need help with the asymptotic normal distribution formula. I could not really figure it out.}$


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