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 p-value?

What conclusion would be made here?

Answer 1

Sample size, n =5 years

Hypothesised mean, =26.30

Population standard deviation, =2.32

Sample mean, =29.94

Since n<30 (small sample), we shall use t-score. For right tailed test, at n - 1 =4 degrees of freedom and at 0.10 alpha, the value of t =1.533

90% one sided confidence interval (right-tailed) for the population mean, is:

​​​​ =(- ​​​​​​, ​​​​​​) =(- ​​​​​​, 29.94 + (1.533*2.32/​​​​​​) =(-​​​​​​, 31.53)

Null Hypothesis:

The average price of tickets has not significantly increased.

Alternative Hypothesis:

The average price of tickets has significantly increased.

Test statistic:

t-stat = =(29.94 - 26.30)/(2.32/​​​​​​) =3.51

Critical value:

Significance level, =0.10

Degrees of freedom, df =n - 1 =4

Type of test =Right-tailed

So, t_critical =1.533

p-value:

p-value for the test statistic of t-stat =3.51 at 4 degrees of freedom for right-tailed test is: p-value =0.012

Conclusion:

Since p-value of 0.012 is less than the significance level(alpha) of 0.10 (0.012 < 0.10), we reject the null hypothesis(H0) at 0.10 significance level.

Thus, there is a sufficient evidence to claim that the average price of tickets has significantly increased.