Question

Please answer each question and explain how you found the correct answer.

2. Additionally, customer satisfaction surveys have indicated that 65% (p=.65) of all TGP customers rate service as good or excellent. You are taking a new survey to see if this is still the case. Assuming a normal distribution, what is the probability that:

a) your inquiry of 90 customers will provide a sample proportion within + and - .10 of the population proportion?

b) Your inquiry of 90 customers will provide a sample proportion between .55 and .70?

3. Suppose you, as the manager of TGP, would like to develop a 95% confidence interval for the number of customers who will enter your store on any given day. Your sample mean was 507 (xbar=507) with a known population standard deviation of 70 customers (o=70). You plan to use a sample size of 50 random days to count customers. Assume a normal distribution.

a) what is the margin of error?

b) What is the confidence interval based on your margin of error?

c) If you wanted to ensure a margin of error of no more than 15 customers, what sample size do you need?

Answer 1

Answer:

giventhat

p0 = 0.65

n = 90

standard error, SE = sqrt(p0(1-p0)/n)

SE = 0.0503

a)

sample proportion within +- .10 of the population proportion

P( 0.55 <P< 0.75 )

=P( ((0.55-0.65) / 0.0503) <Z< ((0.75-0.65) / 0.0503) )

=P( -1.99 <Z< 1.99 )

= P(Z<1.99) - P(Z<-1.99)

= P(Z<1.99) - (1 - P(Z<1.99))

= 0.9767 - ( 1 - 0.9767)

= 0.9767 - 0.0233

= 0.9534

b)

sample proportion between .55 and .70

P( 0.55 <X< 0.7 )

=P( ((0.55-0.65) / 0.0503) <Z< ((0.7-0.65) / 0.0503) )

=P( -1.99 <Z< 0.99 )

= P(Z<0.99) - P(Z<-1.99)

= P(Z<0.99) - (1 - P(Z<1.99))

= 0.8389 - ( 1 - 0.9767)

= 0.8389 - 0.0233

= 0.8156