Question

An educational psychologist wishes to know the mean number of words a third grader can read per minute.
She wants to ensure that the estimate has an error of at most 0.29 words per minute. A previous study found that the mean was 25.7 words per minute.

Assuming that the variance is 8.41, what is the minimum number of third graders that must be included in a sample to construct the 85% confidence interval?

Answer 1

Answer: 208.

Step-by-step explanation:

The formula to find the minimum sample size is given by :-

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

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

$\sigma$ = Standard deviation ( from prior study ) and E = Margin of error.

As per given , we have

Margin of error : E= 0.29

Confidence level = 85%

Significance level =$\alpha=1-0.85=0.15$

Using z-table , the critical value (two -tailed)=$z^*=z_{\alpha/2}=z_{0.15/2}=z_{0.075}=1.439$

As per previous study , Variance =$\sigma^2=8.41$

$\sigma=\sqrt{8.41}=2.9$

Now, the required minimum sample size = $n= (\dfrac{(1.439)\times (2.9)}{0.29})^2$   [Substitute the values in formula (1)]

$n=(14.39)^2$

$n=207.0721\approx208$  [ Round to the next integer]

Hence, the minimum number of third graders that must be included in a sample = 208.