A researcher conducted an Internet survey of 400 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 $225 and the lowest weekly amount being $80. The sample mean is $150. What is the margin of error (ME), rounded to the nearest hundredth, for this college?

The margin of error for 90% confidence interval is 2.98, for 95% confidence interval is 3.55 and for 99% confidence interval is 4.68.

Step-by-step explanation:

The sample size is, n = 400.

The maximum value is, Max. = $225

The minimum value is, Min. = $80

The standard deviation of a distribution using the range of the data is:

As the sample is large, i.e. n = 400 > 30, according to the central limit theorem the sampling distribution of sample mean will follow a normal distribution.

The formula to compute the margin of error is:

         The value of α is 1 - 0.90 = 0.10

         The critical value is,

         (Use the standard normal table)

         The MOE is:

         The value of α is 1 - 0.95 = 0.05

         The critical value is,

         (Use the standard normal table)

         The MOE is:

         The value of α is 1 - 0.99 = 0.01

         The critical value is,

         (Use the standard normal table)

         The MOE is:

Thus, the margin of error for 90% confidence interval is 2.98, for 95% confidence interval is 3.55 and for 99% confidence interval is 4.68.

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step-by-step explanation: