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Chapter 9: Problem 47

Movie Ticket Prices According to Deadline.com, the average price for a movie ticket in 2018 was \(\$ 8.97 .\) A random sample of movie prices in the San Francisco Bay Area 25 movie ticket prices had a sample mean of \(\$ 12.27\) with a standard deviation of \(\$ 3.36\). a. Do we have evidence that the price of a movie ticket in the San Francisco Bay Area is different from the national average? Use a significance level of \(0.05\). b. Construct a \(95 \%\) confidence interval for the price of a movic ticket in the San Francisco Bay Area. How does your confidence interval support your conclusion in part a?

Short Answer

Expert verified
After calculating the Z-score and p-value, we can make a decision about the null hypothesis. If we reject the null hypothesis, that means the movie ticket prices in San Francisco Bay Area are significantly different from the national average. The construction of a 95% confidence interval acts as further evidence supporting the conclusion of our hypothesis test.

Step by step solution

01

Formulate the Hypotheses

The null hypothesis is that the average movie ticket price in the San Francisco Bay Area is the same as the national average, denoted as \(\$8.97\). The alternative hypothesis is that it's different. Using symbolic notation, the hypotheses are: \nH0: 碌 = 8.97 \nH1: 碌 鈮� 8.97
02

Calculate Test Statistic

Next, calculate the z-score. Using the formula \[Z = \((X虅 - 碌) / (蟽/鈭歯)\], where X虅 is the sample mean, 碌 is the population mean, 蟽 is the standard deviation, and n is the sample size. Substituting in the given values, we get: \[Z = \((12.27 - 8.97) / (3.36/ sqrt(25)) \]
03

Check the P-Value and Accept or Reject the Null Hypothesis

Once the Z-score is calculated, find the corresponding p-value. If the p-value is less than the predetermined significance level (0.05), reject the null hypothesis and conclude there is evidence that the local movie ticket price is different from the national average.
04

Construct a 95% Confidence Interval

To construct a 95% confidence interval, use the formula: \[X虅 卤 Z(蟽/鈭歯)\], where Z is the critical value for the desired confidence level, in this case 1.96. Substituting the values, we can calculate the interval.
05

Relate the Confidence Interval to Hypotheses

The 95% confidence interval is the range within which we are fairly sure the true population mean lies with a 95% confidence. If the hypothesized population mean (national average) does not fall within this interval, it stands as additional evidence that the local average price is significantly different from the national average.

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Key Concepts

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

Sample Mean
The sample mean, denoted as \( \bar{x} \), is one of the most fundamental statistics and represents the average value in a sample. It's calculated by summing all the observed values in a sample and dividing by the number of observations. For instance, in the exercise provided, the sample mean of San Francisco Bay Area movie ticket prices is \( \bar{x} = \$12.27 \). This value is crucial when comparing to the population mean to draw conclusions about whether the sample deviates significantly from the population.
Population Mean
The population mean, denoted by \( \mu \), is the average of all measurements in the entire population. In the context of hypothesis testing, it's typically a value that represents the established or hypothesized mean before the research begins. The exercise references the national average movie ticket price as the population mean, \( \mu = \$8.97 \), which serves as a benchmark for comparison against the sample mean.
Standard Deviation
Standard deviation, denoted as \( \sigma \), measures the amount of variation or dispersion of a set of values. A low standard deviation indicates that the values tend to be close to the mean, while a high standard deviation indicates that the values are spread out over a wider range. In the exercise, a standard deviation of \( \sigma = \$3.36 \) for the San Francisco sample implies variability in the movie ticket prices.
Confidence Interval
A confidence interval provides a range of values within which the true population parameter is expected to lie, with a certain degree of confidence. For example, a 95% confidence interval implies that if we were to take 100 different samples and compute a confidence interval for each, we'd expect about 95 of them to contain the actual population mean. The exercise demonstrates constructing a 95% confidence interval to estimate the true mean movie ticket price in the San Francisco Bay Area.
Significance Level
The significance level, denoted as \( \alpha \), is a threshold of probability below which the null hypothesis is rejected in hypothesis testing. A significance level of 0.05, used in the exercise, means there's a 5% risk of concluding that a difference exists when there is no actual difference (a Type I error). It is a way of controlling the probability of making such an error.
Null Hypothesis
The null hypothesis, \( H_0 \), is a default hypothesis that there is no effect or no difference. It is the hypothesis that one aims to test against the alternative hypothesis. In the given exercise, the null hypothesis posits that the average movie ticket price in the San Francisco Bay Area is equal to the national average (\( \mu = \$8.97 \)). It is tested to determine if the observed sample mean statistically significantly differs from \( \mu \) or not.
Alternative Hypothesis
The alternative hypothesis, \( H_1 \), represents what a researcher is trying to prove; it is a statement that indicates a difference or effect. In contrast to the null, the alternative in the exercise is that the mean price of a movie ticket in the San Francisco Bay Area is not equal to the national average (\( \mu eq \$8.97 \)). The evidence collected in the study will support this hypothesis if the null is rejected.
Test Statistic
The test statistic is a standardized value that is calculated from sample data during a hypothesis test. It's used to decide whether to reject the null hypothesis. In the context of the exercise, the test statistic is a z-score that helps determine how many standard deviations the sample mean is from the population mean, accounting for the sample size. The calculated z-score indicates whether the difference between \( \bar{x} \) and \( \mu \) is significant.
P-Value
The p-value is the probability of obtaining test results at least as extreme as the results observed, under the assumption that the null hypothesis is correct. A small p-value, typically less than or equal to the significance level, suggests that the observed data is inconsistent with the null hypothesis. As such, in our exercise, if the p-value is lower than 0.05, we reject the null hypothesis, indicating the average price in San Francisco likely differs from the national average.

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Most popular questions from this chapter

Why Is \(n-1\) in the Sample Standard Deviation? Why do we calculate \(s\) by dividing by \(n-1\), rather than just \(n\) ? $$ s^{2}=\frac{\sum(x-\bar{x})^{2}}{n-1} $$ The reason is that if we divide by \(n-1\), then \(s^{2}\) is an unbiased estimator of \(\sigma^{2}\), the population variance. We want to show that \(s^{2}\) is an unbiased estimator of \(\sigma^{2}\), sigma squared. The mathematical proof that this is true is beyond the scope of an introductory statistics course, but we can use an example to demonstrate that it is. First we will use a very small population that consists only of these three numbers: 1,2, and 5 . You can determine that the population standard deviation, \(\sigma\), for this population is \(1.699673\) (or about \(1.70\) ), as shown in the \(\mathrm{TI}-84\) output. So the population variance, sigma squared, \(\sigma^{2}\), is \(2.888889\) (or about 2.89). Now take all possible samples, with replacement, of size 2 from the population, and find the sample variance, \(s^{2}\), for each sample. This process is started for you in the table. Average these sample variances \(\left(s^{2}\right)\), and you should get approximately \(2.88889 .\) If you do. then you have demonstrated that \(s^{2}\) is an unbiased estimator of \(\sigma^{2}\), sigma squared. Show your work by filling in the accompanying table and show the average of \(s^{2}\).

Exam Scores The distribution of the scores on a certain exam is \(N(100,10)\) which means that the exam scores are Normally distributed with a mean of 100 and a standard deviation of \(10 .\) a. Sketch or use technology to create the curve and label on the \(x\) -axis the position of the mean, the mean plus or minus one standard deviation, the mean plus or minus two standard deviations, and the mean plus or minus three standard deviations. b. Find the probability that a randomly selected score is between 90 and 110 . Shade the region under the Normal curve whose area corresponds to this probability.

Tomatoes Use the data from exercise \(9.36\). a. Using the four-step procedure with a two-sided alternative hypothesis, should you be able to reject the hypothesis that the population mean is 5 pounds using a significance level of \(0.05\) ? Why or why not? The confidence interval is reported here: \(\mathrm{I}\) am \(95 \%\) confident the population mean is between \(4.9\) and \(5.3\) pounds. b. Now test the hypothesis that the population mean is not 5 pounds using the four-step procedure. Use a significance level of \(0.05\) and number your steps.

Ages A study of all the students at a small college showed a mean age of \(20.7\) and a standard deviation of \(2.5\) years. Are these numbers statistics or parameters? Explain. b. Label both numbers with their appropriate symbol (such as \(\bar{x}, \mu, s\), or \(\sigma)\).

Independent or Paired State whether each situation has independent or paired (dependent) samples. a. A researcher wants to compare food prices at two grocery stores. She purchases 20 items at Store A and finds the mean and the standard deviation for the cost of the items. She then purchases 20 items at Store \(\mathrm{B}\) and again finds the mean and the standard deviation for the cost of the items. b. A student wants to compare textbook prices at two bookstores. She has a list of textbooks and finds the price of the text at each of the two bookstores.
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