Question

We are interested in looking at ticket prices of MLB games. It is known from past information that the average price is $26.30, with a population standard deviation of $2.32. Suppose we take a sample of the last 5 years and find that the average ticket price is $29.94. We are interested in seeing if the average price of tickets has significantly increased. Use alpha=.10.

Find the 90% confidence interval for ticket price.

the null and alternative hypotheses are

What is the critical value?

What is the test statistic?

Answer 1

sample mean, xbar = 29.94
sample standard deviation, σ = 2.32
sample size, n = 5

Given CI level is 90%, hence α = 1 - 0.9 = 0.1
α/2 = 0.1/2 = 0.05, Zc = Z(α/2) = 1.64

ME = zc * σ/sqrt(n)
ME = 1.64 * 2.32/sqrt(5)
ME = 1.7

CI = (xbar - Zc * s/sqrt(n) , xbar + Zc * s/sqrt(n))
CI = (29.94 - 1.64 * 2.32/sqrt(5) , 29.94 + 1.64 * 2.32/sqrt(5))

CI = (28.2384 , 31.6416) upto 4 decimal

CI = (28.24 , 31.64) upto 2 decimal

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: μ = 26.3
Alternative Hypothesis, Ha: μ > 26.3

Rejection Region
This is right tailed test, for α = 0.1
Critical value of z is 1.282.
Hence reject H0 if z > 1.282

Test statistic,
z = (xbar - mu)/(sigma/sqrt(n))
z = (29.94 - 26.3)/(2.32/sqrt(5))
z = 3.51