In an effort to estimate the mean of amount spent per customer for dinner at a major lawrence restaurant, data were collected for a sample of 49 customers. assume a population standard deviation of $5. if the sample mean is $24.80, what is the 99% confidence interval for the population mean?

The 99% confidence interval for the population mean is 22.96 to 26.64

Step-by-step explanation:

Consider the provided information,

A sample of 49 customers. Assume a population standard deviation of $5. If the sample mean is $24.80,

The confidence interval if 99%.

Thus, 1-α=0.99

α=0.01

Now we need to determine

Now by using z score table we find that  

The boundaries of the confidence interval are:

Hence, the 99% confidence interval for the population mean is 22.96 to 26.64

a) y = -3x +2

b) y = 1/2x -6

these 2 can be combined into one equation.

-3x +2 = .5x -6

8 = 3.5x

x = 8 / 3.5 = 2.2857142857

y = -3 * (8 / 3.5) +2

y = (-24 / 3.5) + 2

y = -4.8571428571

step-by-step explanation:

the 3rd on is correct

step-by-step explanation: