Question

The chamber of commerce in a tourists resort wishes to estimate the mean expenditure of all tourists who visit the resort. For this purpose a random sample of 100 tourists has been selected for investigation, and it has been found that the sample has a mean expenditure of $8000 and a standard deviation of $900. Find the 95% confidence interval for the mean expenditure of all tourists who visit the resort.

Answer 1

Answer: $(\ 7823.6,\ \ 8176.4)$

Step-by-step explanation:

Given : Sample size : n= 100, it means its a large sample, we use z-test.

Significance level : $\alpha: 1-0.95=0.05$

Critical value : $z_{\alpha/2}=\pm1.96$

Sample mean: $\overline{x}=\ 8000$

Standard deviation: $\sigma=\ 900$

The confidence interval for population mean is given by :-

$\overline{x}\pm z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}\\\\=8000\pm (1.96)\dfrac{900}{\sqrt{100}}\\\\\approx8000\pm176.4=(8000-176.4,8000+176.4)=(7823.6,\ 8176.4)$

Hence, the 95% confidence interval for the mean expenditure of all tourists who visit the resort.= $(\ 7823.6,\ \ 8176.4)$