Question

A retail chain (such as Macy’s) wants to determine the average amount of money which, in
December, customers charge to their Macy’s credit card. A random sample of 100 cards is
selected and sample mean = $730, with sample standard deviation = $200.

a - Build a 95% confidence interval for the mean
b - What should the sample size be if the 95% confidence interval were to have a margin error
of $30?
c - Assume that μ = 720 and σ = 220. When a sample of size 69 is taken, what is the probability
that the sample mean in greater than or equal to 730?
d - Assume that μ = 720 and σ = 220. When a sample of size 64 is taken, what is the
probability that the sample mean in greater than or equal to 730?

Answer 1

a)

95% confidence interval for   is

- Z * / sqrt(n) < < + Z * / sqrt(n)

730 - 1.96 * 200 / sqrt(100) < < 730 + 1.96 * 200 / sqrt(100)

690.8 < < 769.2

95% CI is ( 690.8 , 769.2 )

b)

sample size = ( Z *   / E)²

= ( 1.96 * 200 / 30)²

= 170.74

Sample size = 171 (Rounded up to nearest integer)

c)

Using central limit theorem,

P( ​​​​​​< x) = P( Z < x - / / sqrt(n) )

So,

P( >= 730 ) = P( Z >= 730 - 720 / 220 / sqrt(69) )

= P( Z >= 0.3776)

= 0.3529

b)

P( >= 730 ) = P( Z >= 730 - 720 / 220 / sqrt(64) )

= P(Z >= 0.3636)

= 0.3581