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Chapter 10: Problem 116

Want to attend a pro-basketball finals game? The average prices for the NBA rematch of the Boston Celtics and the LA Lakers in 2010 compared to the average ticket prices in 2008 are given in the table that follows. \({ }^{18}\) $$ \begin{array}{lcc} \text { Game } & \text { 2008 (\$) } & \text { 2010 (\$) } \\ \hline 1 & 593 & 532 \\ 2 & 684 & 855 \\ 3 & 727 & 541 \\ 4 & 907 & 458 \\ 5 & 769 & 621 \\ 6 & 753 & 681 \\ 7 & 533 & 890 \end{array} $$ a. If we were to assume that the prices given in the table have been randomly selected, test for a significant difference between the 2008 and 2010 prices. Use \(\alpha=.01\) b. Find a \(98 \%\) confidence interval for the mean difference, \(\mu_{d}=\mu_{08}-\mu_{10} .\) Does this estimate confirm the results of part a?

Short Answer

Expert verified
Provide evidence to support your answer. Answer: No, there is no significant difference between the average ticket prices for the NBA finals games in 2008 and 2010. The paired t-test with a significance level of 0.01 resulted in a t-statistic of 0.58, which is within the critical value range (卤3.71), leading us to not reject the null hypothesis that there is no significant difference. Additionally, the 98% confidence interval for the mean difference (-246.20 , 357.06) supports this conclusion as it includes both positive and negative values.

Step by step solution

01

Calculate the differences in ticket prices for each game

First, we need to find the differences in ticket prices for each game between 2008 and 2010. Differences (D) = Prices in 2008 - Prices in 2010: $$ D_1 = 593 - 532 = 61 \\ D_2 = 684 - 855 = -171 \\ D_3 = 727 - 541 = 186 \\ D_4 = 907 - 458 = 449 \\ D_5 = 769 - 621 = 148 \\ D_6 = 753 - 681 = 72 \\ D_7 = 533 - 890 = -357 $$ Now we have the differences for each game.
02

Calculate the mean and standard deviation of the differences

Next, we calculate the mean and standard deviation of the differences. Mean of differences: $$\bar{D} = \frac{61 - 171 + 186 + 449 + 148 + 72 - 357}{7} = \frac{388}{7} = 55.43$$ Using the differences, calculate the standard deviation: $$s_D = \sqrt{\frac{(61-55.43)^2+(-171-55.43)^2+(186-55.43)^2+(449-55.43)^2+(148-55.43)^2+(72-55.43)^2+(-357-55.43)^2}{7 - 1}} \approx 242.32$$
03

Calculate the t-statistic and find the critical value for 伪=0.01

We will now calculate the t-statistic for the given paired data using the formula: $$t = \frac{\bar{D}}{s_D/\sqrt{n}} = \frac{55.43}{242.32/\sqrt{7}} \approx 0.58$$ Also, find the critical value for 伪=0.01 using a t-distribution table. The degrees of freedom (df) for this test are n-1 = 7-1 = 6. The critical value for a two-tailed test at 伪=0.01 and df=6 is approximately 卤3.71.
04

Test the t-statistic against the critical value and make a conclusion

Compare the calculated t-statistic (0.58) to the critical value (卤3.71). Since the t-statistic is within the critical value range, we cannot reject the null hypothesis that there is no significant difference between the average ticket prices in 2008 and 2010.
05

Calculate the 98% confidence interval for the mean difference

The 98% confidence interval for the mean difference (渭_d) can be calculated using the formula: $$\bar{D} \pm t_{critical} \cdot \frac{s_D}{\sqrt{n}}$$ Using the critical value (3.71) from Step 3, the confidence interval is: $$55.43 \pm 3.71 \cdot \frac{242.32}{\sqrt{7}} \approx (-246.20, 357.06)$$
06

Interpret the confidence interval

The 98% confidence interval suggests that with 98% confidence, the true mean difference between 2008 and 2010 average ticket prices is between -\(246.20 and \)357.06. Given that the confidence interval includes both positive and negative values, it supports the conclusion from part a that there is no significant difference between the average ticket prices in 2008 and 2010.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Paired t-test
A paired t-test is a statistical method used to compare two related groups. It's handy when you have paired data like before and after measurements鈥攕uch as ticket prices from 2008 and 2010 in this problem. The paired t-test accounts for the fact that the data points in each pair may be correlated.

To perform a paired t-test, you calculate the differences between the paired values (e.g., price in 2008 minus price in 2010), and then evaluate whether the average of these differences is significantly different from zero.
  • First, find the difference for each pair.
  • Calculate the mean of these differences.
  • Then find the standard deviation.
  • Finally, use these values to compute a t-statistic.

If your calculated t-value exceeds the critical value from the t-distribution table, you reject the null hypothesis that there is no difference between the two sets of data. In this scenario, because the t-value did not exceed the critical value, we conclude there is no significant difference between ticket prices of 2008 and 2010.
Confidence Interval
A confidence interval offers a range of values, derived from sample data, likely to contain the true population parameter. In our exercise, the confidence interval estimates the mean difference in ticket prices between 2008 and 2010.

Constructing a confidence interval involves some important steps:
  • Start with the mean difference from your paired data.
  • Utilize the t-score corresponding to your desired confidence level (98% in this case).
  • Calculate the margin of error using the t-score, standard deviation of the differences, and the square root of the number of pairs (n).
Adding and subtracting the margin of error from the mean gives you the lower and upper bounds of the confidence interval.

The confidence interval in this exercise spans from -246.20 to 357.06. Because this range includes zero, it implies no clear difference in the average prices, echoing the paired t-test results.
Significance Level
The significance level, often denoted by 伪, represents the threshold for determining whether a result is statistically significant. In hypothesis testing, it defines the probability of rejecting the null hypothesis when it is actually true.
  • A common significance level is 0.05; however, in this exercise, we use an 伪 of 0.01, indicating a more stringent criterion.
  • This means we're willing to accept a 1% chance of committing a Type I error, which is falsely concluding that the average ticket prices differ.

The choice of 伪 affects the critical value from the t-distribution table.
  • A smaller 伪 (like 0.01) results in a larger critical value, making it harder to claim significance.
  • In our problem, with df=6, the critical value was 卤3.71.
The calculated t-statistic of 0.58 lies well within this critical range, confirming no significant price difference. This reinforces our comprehensive understanding from both the confidence interval and the paired t-test results.

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Most popular questions from this chapter

Organic chemists often purify organic compounds by a method known as fractional crystallization. One chemist prepared ten \(4.85-\mathrm{g}\) quantities of aniline and purified it to acetanilide. The following dry yields were recorded: $$ \begin{array}{lllllll} 3.85 & 3.80 & 3.88 & 3.85 & 3.90 \\ 3.36 & 3.62 & 4.01 & 3.72 & 3.83 \end{array} $$ Estimate the mean grams of acetanilide that can be recovered from an initial amount of \(4.85 \mathrm{~g}\) of aniline. Use a \(95 \%\) confidence interval.

An article in American Demographics investigated consumer habits at the mall. We tend to spend the most money shopping on the weekends, and, in particular, on Sundays from 4 to 6 P.M. Wednesday morning shoppers spend the least! \({ }^{19}\) Suppose that a random sample of 20 weekend shoppers and a random sample of 20 weekday shoppers were selected, and the amount spent per trip to the mall was recorded. $$ \begin{array}{lcc} & \text { Weekends } & \text { Weekdays } \\ \hline \text { Sample Size } & 20 & 20 \\ \text { Sample Mean (\$) } & 78 & 67 \\ \text { Sample Standard Deviation (\$) } & 22 & 20 \end{array} $$ a. Is it reasonable to assume that the two population variances are equal? Use the \(F\) -test to test this hypothesis with \(\alpha=.05\). b. Based on the results of part a, use the appropriate test to determine whether there is a difference in the average amount spent per trip on weekends versus weekdays. Use \(\alpha=.05\).

Find the critical value(s) of \(t\) that specify the rejection region in these situations: a. A two-tailed test with \(\alpha=.01\) and \(12 d f\) b. A right-tailed test with \(\alpha=.05\) and \(16 d f\) c. A two-tailed test with \(\alpha=.05\) and \(25 d f\) d. A left-tailed test with \(\alpha=.01\) and \(7 d f\)

An experiment was conducted to compare the densities (in ounces per cubic inch) of cakes prepared from two different cake mixes. Six cake pans were filled with batter \(\mathrm{A}\), and six were filled with batter B. Expecting a variation in oven temperature, the experimenter placed a pan filled with batter \(\mathrm{A}\) and another with batter \(\mathrm{B}\) side by side at six different locations in the oven. The six paired observations of densities are as follows: $$ \begin{array}{lrrrrrr} \text { Location } & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline \text { Batter A } & .135 & .102 & .098 & .141 & .131 & .144 \\ \hline \text { Batter B } & .129 & .120 & .112 & .152 & .135 & .163 \end{array} $$ a. Do the data present sufficient evidence to indicate a difference between the average densities of cakes prepared using the two types of batter? b. Construct a \(95 \%\) confidence interval for the difference between the average densities for the two mixes.

An experiment published in The American Biology Teacher studied the efficacy of using \(95 \%\) ethanol or \(20 \%\) bleach as a disinfectant in removing bacterial and fungal contamination when culturing plant tissues. The experiment was repeated 15 times with each disinfectant, using eggplant as the plant tissue being cultured. \({ }^{8}\) Five cuttings per plant were placed on a petri dish for each disinfectant and stored at \(25^{\circ} \mathrm{C}\) for 4 weeks. The observation reported was the number of uncontaminated eggplant cuttings after the 4 -week storage. $$ \begin{array}{lcl} \text { Disinfectant } & 95 \% \text { Ethanol } & 20 \% \text { Bleach } \\ \hline \text { Mean } & 3.73 & 4.80 \\ \text { Variance } & 2.78095 & .17143 \\ n & 15 & 15 \end{array} $$ a. Are you willing to assume that the underlying variances are equal? b. Using the information from part a, are you willing to conclude that there is a significant difference in the mean numbers of uncontaminated eggplants for the two disinfectants tested?
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