PLEASE HELP WILL MARK BRAINLIEST You work for a consumer advocate agency and want to find the mean repair cost of a washing machine. As part of your study, you randomly select 40
repair costs and find the mean to be $120.00. The sample standard deviation
is $17.50. The 99% confidence interval for the population mean repair cost is? A.(112.86, 127.14) B.(114.58, 125.42) C. (115.43, 124.57) D. (111.57, 128.43)

The correct answer is - A.(112.86, 127.14)

Step-by-step explanation:

Given:

mean = $120

sd = $17.50

n = 40

Solution:

Confidence interval for a populationcan be express as mean +/- margin of error (E)

degree of freedom = n-1 = 40-1 = 39

confidence level (C) = 99% = 0.01

significance level = 1 - C = 1 - 0.01 = 0.99 = 99%

 by margin of error E = t×sd/√n = 2.58*17.50/√40

= 2.58*2.76

= 7.138 or 7.14

then the lower limit of mean = mean - E = 120 - 7.14 = $112.86

and, the upper limit of population mean = mean + E = 120 + 7.14 = $127.14

15 batches are the most you can make

step-by-step explanation: