Question

2. Can you afford the price of an NBA ticket during the regular season? The Website www.answers.com indicates that the low prices are around $10 for the high up seats while the court-side seats are around $2000 to $5000 per game and the average price of a ticket is $75.5 a game. Suppose that we test this claim by selecting a random sample of = 50 ticket purchases from a computer database and find that the average ticket price is $82.50 with a standard deviation of $75.25.

a.Do you think that x, the price of an individual regular season ticket, has a mound-shaped distribution? If not, what shape would you expect?
b.If the distribution of the ticket prices is not normal, you can still use the standard normal distribution to construct a confidence interval for , the average price of a ticket. Why?
c. Construct a 95% confidence interval for , the average price of a ticket. Does this confidence interval cause you support or question the claimed average price of $75.50? Explain. Show all five steps.
d. Suppose that a sports analyst wants a better estimate of , the average price of a ticket. They made a rough guess of 72.5 for the value of the standard deviation before collecting data. What sample size would be necessary to obtain an interval width of 30 for a confidence level of 95%? Show work.

Answer 1

a. The shape of the ticket price distribution is likely skewed to the right. b. The standard normal distribution can still be used to construct a confidence interval for the average price of a ticket. c. The 95% confidence interval supports the claimed average price of $75.50.

Step-by-step explanation:

a. A mound-shaped distribution, also known as a normal distribution, is symmetrical around its mean. In this case, we don't know the shape of the distribution because we don't have information about skewness or kurtosis. However, since price cannot be negative and there's a possibility of very high prices, we would expect the distribution to be skewed to the right.

b. If the distribution of ticket prices is not normal, we can still use the standard normal distribution to construct a confidence interval for the average price of a ticket. This is because of the Central Limit Theorem, which states that as sample size increases, the sampling distribution of the sample mean approaches a normal distribution, regardless of the shape of the population distribution.

c. To construct a 95% confidence interval for the average price of a ticket, we can use the formula: CI = x +/- (Z * (s/√n)), where x is the sample mean, Z is the Z-score corresponding to the desired confidence level, s is the sample standard deviation, and n is the sample size. Plugging in the values from the question, we get: CI = 82.5 +/- (1.96 * (75.25/√50)). Calculating this gives us a confidence interval of approximately $67.78 to $97.22. Since the claimed average price of $75.50 falls within this confidence interval, it supports the claimed average price.

d. To find the necessary sample size to obtain an interval width of 30 for a confidence level of 95%, we can use the formula: n = (Z * s / E)^2, where Z is the Z-score corresponding to the desired confidence level, s is the estimated standard deviation, and E is the desired margin of error. Plugging in the values from the question, we get: n = (1.96 * 72.5 / 30)^2. Calculating this gives us a necessary sample size of approximately 39.