Question

In a group of 14 young professionals studied, it was found that they purchased on average $171.719 per month eating meals prepared outside the home, with a standard deviation of $36.011. What is the 90% confidence interval of the true amount of money young professionals spend monthly on meals prepared outside the home?

Question 3 options:

	1) ( 162.095 , 181.343 )
	2) ( 169.948 , 173.49 )
	3) ( -154.675 , 188.763 )
	4) ( 154.675 , 188.763 )
	5) ( 154.771 , 188.667 )

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 171.719

sample standard deviation = s = 36.011

sample size = n = 14

Degrees of freedom = df = n - 1 = 13

At 90% confidence level the t is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

t _/2,df = t_0.05,13 = 1.771

Margin of error = E = t_/2,df * (s /n)

= 1.771* (36.011 / 14)

= 17.045

The 90% confidence interval estimate of the population mean is,

- E < < + E

171.719 - 17.045 < < 171.719 + 17.045

154.675 < < 188.763

(154.675 , 188.763)

Option 4) is correct.