Question

Suppose we want a 90% confidence interval for the average amount of time (in minutes) spent per week on homework by the students in a large introductory statistics course at a large university. The interval is to have a margin of error of 3 minutes, and the amount spent has a Normal distribution with a standard deviation σ = 40 minutes. The number of observations required is closest to:
1180.
683.
482.
22.

Answer 1

Answer: 482

Step-by-step explanation:

Formula to find the sample size is given by :-

$n= (\dfrac{z^*\times \sigma}{E})^2$                                    (1)

, where z* = critical z-value (two tailed).  

$\sigma$ = Population standard deviation and E = Margin of error.

As per given , we have

Margin of error : E= 3

$\sigma=40$

Confidence level = 90%

Significance level =$\alpha=1-0.90=0.10$

Using z-table , the critical value for 90% confidence=$z^*=z_{\alpha/2}=z_{0.05}=1.645$

Required minimum sample size = $n= (\dfrac{(1.645)\times (40)}{3})^2$   [Substitute the values in formula (1)]

$n=(21.9333333333)^2$

$n=481.07111111\approx482$  [ Round to the next integer]

Hence, the number of observations required is closest to 482.