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Finger [1]
3 years ago
10

A researcher conducted an Internet survey of 500 students at a particular college to estimate the average amount of money studen

ts 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?
Mathematics
1 answer:
bonufazy [111]3 years ago
6 0

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

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