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)`

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)(\sigma)/(√(n))\\\\=8000\pm (1.96)(900)/(√(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)`