Question

Suppose the number of dollars spent per week on groceries is normally distributed. If the population standard deviation is 7 dollars, what minimum sample size is needed to be 90% confident that the sample mean is within 3 dollars of the true population mean?

Answer 1

Answer: 15

Explanation:

Given : Level of confidence = 0.90

Significance level :
`\alpha=1-0.90=0.10`

Critical value :
`z_(\alpha/2)=1.645`

Margin of error :
`E=\text{ 3 dollars}`

Standard deviation:
`\sigma=\text{ 7 dollars}`

The formula to find the sample size :
`n=((\sigma* z_(\alpha/2))/(E))^2`

`\Rightarrow n=((7*(1.645))/(3))^2=14.7328027778\approx15`

Hence, the minimum sample size needed= 15.