Question

In a group of 12 young professionals studied, it was found that they purchased on average $245.804 per month eating meals prepared outside the home, with a standard deviation of $34.4307. What is the 95% confidence interval of the true amount of money young professionals spend monthly on meals prepared outside the home?

Question 1 options:

	1) ( 224.146 , 267.462 )
	2) ( 223.928 , 267.68 )
	3) ( -223.928 , 267.68 )
	4) ( 235.865 , 255.743 )
	5) ( 243.603 , 248.005 )

Suppose you work for Fender Guitar Company and you are responsible for testing the integrity of a new formulation of guitar strings. To perform your analysis, you randomly select 55 'high E' strings and put them into a machine that simulates string plucking thousands of times per minute. You record the number of plucks each string takes before failure and compile a dataset. You find that the average number of plucks is 6,420.7 with a standard deviation of 255.01. A 99% confidence interval for the average number of plucks to failure is (6,328.9, 6,512.5). From the option listed below, what is the appropriate interpretation of this interval?

Question 2 options:

	1) We are 99% confident that the average number of plucks to failure for all 'high E' strings is between 6,328.9 and 6,512.5.
	2) We are certain that 99% of the average number of plucks to failure for all 'high E' strings will be between 6,328.9 and 6,512.5.
	3) We are 99% confident that the average number of plucks to failure for all 'high E' strings tested is between 6,328.9 and 6,512.5.
	4) We are 99% confident that the proportion of all 'high E' guitar strings fail with a rate between 6,328.9 and 6,512.5.
	5) We cannot determine the proper interpretation of this interval.

To design a new advertising campaign, Ford Motor Company would like to estimate the proportion of drivers of the new Ford Fusion that are women. In a random sample of 95 Fusion owners, 45 of them were women. What is the 90% confidence interval estimating the proportion of all drivers who are women?

Question 3 options:

	1) ( 0.44205 , 0.61058 )
	2) ( -0.38942 , 0.55795 )
	3) ( 0.42246 , 0.52491 )
	4) ( 0.40803 , 0.53934 )
	5) ( 0.38942 , 0.55795 )

Answer 1

Solution :

Given that,

1) Point estimate = sample mean = = 245.804

sample standard deviation = s = 34.4307

sample size = n = 12

Degrees of freedom = df = n - 1 = 12-1 = 11

At 95% confidence level

= 1-0.95% =1-0.95 =0.05

/2 =0.05/ 2= 0.025

t/2,df = t0.025,11 = 2.2

t /2,df = 2.2

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

= 2.2 * (34.4307 / 12)

Margin of error = E = 21.876

The 95% confidence interval estimate of the population mean is,

- E < <  + E

245.804 - 21.876 < < 245.804 + 21.876

223.928 < < 267.68

(223.928,267.68)

Answer = 2) (223.928,267.68)

2) Answer =

1) We are 99% confident that the average number of plucks to failure for all 'high E' strings is between 6,328.9 and 6,512.

3)

n = 95

x = 45

Point estimate = sample proportion = = x / n = 45/95 = 0.474

1 - = 1 - 0.474 = 0.526

At 90% confidence level

= 1-0.90% =1-0.90 =0.10

/2 =0.10/ 2= 0.05

Z/2 = Z0.05 = 1.645

Z/2 = 1.645  

Margin of error = E = Z / 2 * (( * (1 - )) / n)

= 1.645 * ((0.474*(0.526) /95 )

= 0.08458

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.474 - 0.08458 < p < 0.474 + 0.08458

0.38942 < p < 0.55795

( 0.38942 , 0.55795 )

ANSWER = 5)( 0.38942 , 0.55795 )