Question

A researcher is planning to compare the effects of two different types of lights on the growth of bean plants. She expects that the means of the two groups will differ by about 1 inch and that in each group the standard deviation of plant growth will be around 1.5 inches. Consider the guideline that the anticipated SE for each experimental group should be no more than one-fourth of the anticipated difference between the two group means. How large should the sample be (for each group) in order to meet this guideline

Answer 1

Answer:

The sample size is  $n = 36$

Step-by-step explanation:

From the question we are told that

   The sample standard deviation is  $s = 1.5$

  The mean difference of the two groups is  $\=x _ 1 - \=x_2 = 1$

   The standard error is  $SE = \frac{ \= x_1 - \= x_2}{4}$

=>                                      $SE = \frac{1}{4}$

Let assume that the confidence level is 95% hence the level of significance is  

   $\alpha = (100-95)\%$

=> $\alpha = 0.05$

So the critical value of  $\frac{\alpha }{2}$ obtained from the normal distribution table is

     $Z_{\frac{\alpha }{2} } = 1.96$

Generally the margin of error is mathematically evaluated as

     $E = Z_{\frac{\alpha }{2} } * SE$    

   $E = 1.96 * \frac{1}{4}$

    $E = 0.49$

Generally the sample size is mathematically represented as

  $n =[ \frac{Z_{\frac{\alpha }{2} * s^2}}{E}]^2$

=> $n = [\frac{1.96 * 1.5}{0.49} ]^2$

=> $n = 36$