Question

A researcher conducted an Internet survey of 500 students at a particular college to estimate the average amount of money students spend on groceries per week. The data are normally distributed, with the highest weekly amount being $200 and the lowest weekly amount being $75. The sample mean is $140. What is the margin of error (ME), rounded to the nearest hundredth, for this college?

Answer 1

Answer:

The margin of error is  $ME = 2.2962$

Step-by-step explanation:

From the question we are told that

     The number of student is $n = 500$

     The highest amount is  $A =$$200

     The lowest amount is  $B =$ $75

       The sample mean is  $x =$ &140

The Standard deviation  of this set is mathematically evaluated as

            $s = \frac{A-B}{4}$

Substituting values

            $s = \frac{200-75}{4}$

            $s = 31.25$

The margin of error (ME) is mathematically evaluated as

            $ME = t_{n-1}__{\alpha }} * \frac{s}{\sqrt{n} }$

Where  $t_{n-1}__{\alpha }}$ is the critical value for $\alpha = 0.05$ i.e the significance level

     From the critical value table this is  $t_{n-1}__{\alpha }} = 1.649$

So

       $ME = 1.649 * \frac{31.25}{\sqrt{500} }$

      $ME = 2.2962$