Question

157 employees of your firm were asked about their job satisfaction. Out of the 157, 78 said they were unsatisfied. What is the estimate of the population proportion? What is the standard error of this estimate?

Question 1 options:

	1) Estimate of proportion: 0.497, Standard error: 0.0032.
	2) Estimate of proportion: 0.503, Standard error: 0.0032.
	3) Estimate of proportion: 0.503, Standard error: 0.0399.
	4) Estimate of proportion: 0.497, Standard error: 0.0399.
	5) The true population proportion is needed to calculate this.

Suppose the nationwide high school dropout rate for 2014 was around 14.62%. If you checked the records of 472 students in high school in 2014, what is the probability that greater than 14.97% of them ended up dropping out?

Question 2 options:

	1) 0.5852
	2) -22.9702
	3) <0.0001
	4) 0.5000
	5) 0.4148

Question 3 (1 point)

Fill in the blank. In a drive thru performance study, the average service time for McDonald's is 200.64 seconds with a standard deviation of 8.89 seconds. A random sample of 84 times is taken. There is a 19% chance that the average drive-thru service time is less than ________ seconds.

Question 3 options:

	1) 192.84
	2) 199.79
	3) There is not enough information to determine this.
	4) 201.49
	5) 208.44

Answer 1

1)

sample proportion     p̂		x/n=		0.497
std error       =Se		=√(p*(1-p)/n) =		0.0399

option 4 is correct

2)

for normal distribution z score =(p̂-p)/σp
here population proportion=     p=	0.1462
sample size       =n=	472
std error of proportion=σp=√(p*(1-p)/n)=	0.0163

probability =

P(X>0.1497)

=

P(Z>0.22)=

1-P(Z<0.22)=

1-0.5852=

0.4148

option 5 is correct

3)

mean μ=	200.640
standard deviation σ=	8.890
sample size       =n=	84
std error=σx̅=σ/√n=	0.9700

for 19th percentile critical value of z=		-0.88
therefore corresponding value=mean+z*std deviation=			199.79

option 2 is correct