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

The margin of error is
`ME = 2.2962`

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 = (A-B)/(4)`

Substituting values

`s = (200-75)/(4)`

`s = 31.25`

The margin of error (ME) is mathematically evaluated as

`ME = t_(n-1)__(\alpha )} * (s)/(√(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 * (31.25)/(√(500) )`

`ME = 2.2962`