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.

idk

step-by-step explanation:

use that smart brain

check above link.

graphically?   well, you graph them both, the solutions is where they both intersect each other, since there are two solutions, meaning they will meet twice, or namely the linear will run into the quadratic curve twice, and where they meet, those x,y coordinates are the solutions to the system.