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Chapter 9: Problem 41

The XO Group Inc., released the results of its annual Real Weddings Study on March 27, 2014 (www.theknot.com). According to this study, the average cost of a wedding in America was \(\$ 29,858\) in \(2013 .\) A recent sample of 100 American couples who got married this year produced a mean wedding cost of \(\$ 32,084\) with a standard deviation of \(\$ 9275 .\) Using a \(2.5 \%\) significance level and the criticalvalue approach, can you conclude that the current mean cost of a wedding in America is higher than \(\$ 29,858\) ? Find the range for the \(p\) -value for this test. What will your conclusion be using this \(p\) -value range and \(\alpha=.05 ?\)

Short Answer

Expert verified
The answer will depend on the results of the calculations. If the calculated t-value is greater than the critical t-value, and the p-value (given its range) is less than 0.05, then we would reject the null hypothesis and conclude that the current mean cost of a wedding in America is higher than $29,858. If not, we cannot conclude that.

Step by step solution

01

State the hypotheses

The null hypothesis is that the current mean wedding cost is equal to $29,858 (or, less than or equal to $29,858). The alternative hypothesis is that the current mean wedding cost is greater than $29,858. In mathematical terms, this can be represented as: \(H_0: \mu \leq 29,858\) and \(H_a: \mu > 29,858\).
02

Compute the test statistic

To test the hypothesis, compute the one-sample t-statistic, follows the formula: \( t = \frac{x - \mu }{(s/ \sqrt{n})} \) where \(x\) represents the sample mean, \(\mu\) represents the population mean, \(s\) represents the sample standard deviation, and \(n\) indicates the sample size. Hence, substituting the values, the t-value comes out to be \( t = \frac{32,084 - 29,858}{(9275 / \sqrt{100})} \).
03

Determine the critical value

Since the significance level, \(\alpha\), is given as 0.025 (2.5% converted to decimal), and it's a one-tailed test (as we are testing for greater than), using the t-distribution table or a calculator, the critical t-value for df = 99 (n-1 = 100-1) at a 0.025 significance level is approximately 1.984.
04

Make the statistical decision

If the computed t-value is greater than the critical t-value, then we reject the null hypothesis. This means we can conclude that the current mean cost of a wedding in America is higher than $29,858.
05

Find the range for the p-value

The p-value is the smallest level of significance at which the null hypothesis would be rejected. It can be obtained from the t-distribution table or through a statistical software by considering the area in the tail (right tail, since it's a greater than test) given the computed t-value. The p-value range requested is likely the range between the significance level at which the null hypothesis is not rejected, and the significance level at which the null hypothesis is rejected.
06

Make the conclusion for \(\alpha=.05\)

If the p-value is less than 0.05, we reject the null hypothesis and conclude that the current mean cost of a wedding in America is higher than $29,858. If the p-value is greater than 0.05, we fail to reject the null hypothesis, and hence, we do not have enough evidence to conclude that the current mean cost of a wedding in America is higher than $29,858.

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Key Concepts

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

t-distribution
The t-distribution is an essential concept in statistics, particularly when dealing with smaller sample sizes or unknown population variances.
When a sample size is small, or the population standard deviation is unknown, the t-distribution provides a more accurate update than the normal distribution.
In this problem, because the sample size is 100 and the population standard deviation is not provided, we use t-distribution.
  • The t-distribution is often used when the sample size is less than 30, but it's still relevant when larger samples are drawn from normally distributed populations without a known standard deviation.
  • The shape of the t-distribution resembles a normal curve but with fatter tails, accommodating more variability and accounting for the additional uncertainty.
  • As the sample size increases, the t-distribution approaches the normal distribution.
In this exercise, using the t-distribution is crucial to compute the test statistic that correctly reflects the sample variability.
significance level
The significance level, often denoted as \(\alpha\), is a threshold set by the researcher to determine when to reject the null hypothesis.
It plays a vital role in hypothesis testing, providing a benchmark for the probability of observing results at least as extreme as the ones obtained, assuming the null hypothesis is true.
  • The significance level indicates the likelihood of making a Type I error, which occurs when the true null hypothesis is incorrectly rejected.
  • A common choice for the significance level is 0.05, but this problem uses a more stringent 0.025 or 2.5% threshold.
  • If the p-value is less than the significance level, the results are statistically significant, leading to the rejection of the null hypothesis.
This test's significance level of 0.025 implies that the researcher is willing to accept a 2.5% chance of incorrectly claiming that the mean wedding cost is higher than \(\$29,858\). This threshold directly influences the determination of the critical t-value.
null hypothesis
The null hypothesis is a statement in hypothesis testing that there is no effect or difference in the population against which the alternative hypothesis is tested.
It serves as the initial claim that researchers seek to test.
  • In mathematical terms, the null hypothesis for this exercise states: \(H_0: \mu \leq 29,858\). This suggests that the current mean cost of a wedding is less than or equal to \(\$29,858\).
  • Testing the null hypothesis involves using sample data to determine whether there is enough statistical evidence to reject it in favor of the alternative hypothesis.
  • The alternative hypothesis in this case states \(H_a: \mu > 29,858\), indicating a belief that the mean wedding cost has indeed increased.
Understanding the null hypothesis is crucial, as the entire hypothesis testing framework hinges upon whether or not we reject this initial assertion.
p-value
The p-value is a measure used in hypothesis testing to determine the strength of evidence against the null hypothesis.
It represents the probability of observing the test statistic, or something more extreme, assuming the null hypothesis is true.
  • A low p-value indicates strong evidence against the null hypothesis, suggesting that such extreme observations are unlikely under it.
  • If the p-value is less than or equal to the chosen significance level (\(\alpha\)), the null hypothesis is rejected.
  • In contrast, a high p-value suggests that the data is consistent with the null hypothesis.
In this exercise, finding the p-value involves calculating the area in the tail of the t-distribution corresponding to the calculated t-statistic.
The provided solution asks us to compare the p-value range to a significance level of 0.05 to ascertain whether or not to reject the null hypothesis of the wedding cost.

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Most popular questions from this chapter

Lazurus Steel Corporation produces iron rods that are supposed to be 36 inches long. The machine that makes these rods does not produce each rod exactly 36 inches long. The lengths of the rods are approximately normally distributed and vary slightly. It is known that when the machine is working properly, the mean length of the rods is 36 inches. The standard deviation of the lengths of all rods produced on this machine is always equal to \(.035\) inch. The quality control department at the company takes a sample of 20 such rods every week, calculates the mean length of these rods, and tests the null hypothesis, \(\mu=36\) inches, against the alternative hypothesis, \(\mu \neq 36\) inches. If the null hypothesis is rejected, the machine is stopped and adjusted. A recent sample of 20 rods produced a mean length of \(36.015\) inches. a. Calculate the \(p\) -value for this test of hypothesis. Based on this \(p\) -value, will the quality control inspector decide to stop the machine and adjust it if he chooses the maximum probability of a Type I error to be .02? What if the maximum probability of a Type I error is \(10 ?\) b. Test the hypothesis of part a using the critical-value approach and \(\alpha=.02\). Does the machine need to be adjusted? What if \(\alpha=.10 ?\)

Find the \(p\) -value for each of the following hypothesis tests. a. \(H_{0}: \mu=23, \quad H_{1}: \mu \neq 23, \quad n=50, \quad \bar{x}=21.25, \quad \sigma=5\) b. \(H_{0}: \mu=15, \quad H_{1}: \mu<15, \quad n=80, \quad \bar{x}=13.25, \quad \sigma=5.5\) c. \(H_{0}: \mu=38, \quad H_{1}: \mu>38, \quad n=35, \quad \bar{x}=40.25, \quad \sigma=7.2\)

Customers often complain about long waiting times at restaurants before the food is served. A restaurant claims that it serves food to its customers, on average, within 15 minutes after the order is placed. \(A\) local newspaper journalist wanted to check if the restaurant's claim is true. A sample of 36 customers showed that the mean time taken to serve food to them was \(15.75\) minutes with a standard deviation of \(2.4\) minutes. Using the sample mean, the joumalist says that the restaurant's claim is false. Do you think the journalist's conclusion is fair to the restaurant? Use a \(1 \%\) significance level to answer this question.

Briefly explain the meaning of each of the following terms. a. Null hypothesis b. Alternative hypothesis c. Critical point(s) d. Significance level e. Nonrejection region f. Rejection region \(\mathrm{g}\). Tails of a test h. Two types of errors

Consider the following null and alternative hypotheses: $$ H_{0}: p=.82 \text { versus } H_{1}: p \neq .82 $$ A random sample of 600 observations taken from this population produced a sample proportion of \(.86\). a. If this test is made at a \(2 \%\) significance level, would you reject the null hypothesis? Use the critical-value approach. b. What is the probability of making a Type I error in part a? c. Calculate the \(p\) -value for the test. Based on this \(p\) -value, would you reject the null hypothesis if \(\alpha=.025\) ? What if \(\alpha=.01 ?\)
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