/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 40 The cost of weddings in the Unit... [FREE SOLUTION] | 91Ó°ÊÓ

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The cost of weddings in the United States has skyrocketed in recent years. As a result many couples are opting to have their weddings in the Caribbean. A Caribbean vacation resort recently advertised in Bride Magazine that the cost of a Caribbean wedding was less than \(\$ 10,000\). Listed below is a total cost in \(\$ 000\) for a sample of 8 Caribbean weddings. $$ \begin{array}{llllllll}9.7 & 9.4 & 11.7 & 9.0 & 9.1 & 10.5 & 9.1 & 9.8 \\\\\hline\end{array}$$ At the .05 significance level is it reasonable to conclude the mean wedding cost is less than \(\$ 10,000\) as advertised?

Short Answer

Expert verified
It's not reasonable to conclude the mean cost is less than $10,000.

Step by step solution

01

State the Hypotheses

To determine whether the mean cost is less than \(10,000, we set up the null hypothesis (H_0) and the alternative hypothesis (H_a). The null hypothesis is H_0: \mu = 10 and the alternative hypothesis is H_a: \mu < 10, where \mu represents the mean cost of a Caribbean wedding in \)000.
02

Calculate the Sample Mean and Sample Standard Deviation

First, calculate the sample mean \( \bar{x} \) using the formula:\[ \bar{x} = \frac{\sum x_i}{n} \]where \( x_i \) are the data values and \( n = 8 \) is the sample size.\[ \bar{x} = \frac{9.7 + 9.4 + 11.7 + 9.0 + 9.1 + 10.5 + 9.1 + 9.8}{8} = 9.9125 \]Next, calculate the sample standard deviation \( s \) using:\[ s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}} \]
03

Compute the Test Statistic

Use the t-statistic formula:\[ t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} \]where \( \mu_0 = 10 \), \( \bar{x} = 9.9125 \), \( s \approx 0.861 \) (after calculation), and \( n = 8 \).\[ t = \frac{9.9125 - 10}{0.861/\sqrt{8}} \approx -0.287 \]
04

Determine the Critical Value and Decision Rule

For a one-tailed test at a 0.05 significance level with degrees of freedom ( = n-1 = 7 ), find the critical t-value from the t-distribution table. The critical t-value is approximately -1.895. If our calculated t-statistic is less than this critical t-value, we will reject the null hypothesis.
05

Make a Decision

Since the calculated t-statistic \(-0.287\) is greater than the critical value \(-1.895\), we fail to reject the null hypothesis.

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Key Concepts

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

t-distribution
In hypothesis testing, especially with small sample sizes (typically under 30), we often rely on the t-distribution rather than the normal distribution. This is because the t-distribution takes into account the increased variability found in smaller samples, providing a more accurate result.
The t-distribution is similar in shape to the normal distribution but has heavier tails. It accounts for the uncertainty in the estimate of the standard deviation when working with small samples. As sample size increases, the t-distribution approaches the normal distribution.
A key component of using the t-distribution is understanding the degrees of freedom, which typically equals the sample size minus one \(df = n - 1\). In our wedding cost sample, with 8 weddings, the degrees of freedom is 7. This affects the critical value we use in our hypothesis test.
sample mean
The sample mean is crucial in hypothesis testing as it is our best estimate of the population mean. It provides a central value around which the data points are distributed.
To calculate the sample mean, sum all the data values and divide by the number of observations. In our Caribbean weddings example, the total of the sample data equates to 79.3 when added together. By dividing by the number of weddings (8), we get a sample mean \( \bar{x} = 9.9125 \).
The sample mean serves as the starting point in hypothesis tests, evaluating differences between the sample mean and the specified value (here, $10,000), thereby assisting in making inferences about the population mean.
significance level
The significance level, often denoted by \( \alpha \), is the probability of rejecting the null hypothesis when it is true. It represents the risk we are willing to take for a type I error. Common significance levels include 0.05, 0.01, and 0.10.
In our test about the cost of Caribbean weddings, we use a significance level of 0.05. This implies there is a 5% risk of concluding that the mean wedding cost is less than \(10,000 when it might actually not be. At this level, the decision rule involves comparing the calculated t-statistic to the critical t-value found in the t-distribution table with 7 degrees of freedom.
This critical value acts as a threshold determining if the observed data sufficiently supports rejecting the null hypothesis. Since the calculated t-statistic was not less than the critical t-value, we did not find enough evidence to support the alternative hypothesis that the cost is below \)10,000.
one-tailed test
A one-tailed test in hypothesis testing checks whether a sample mean is either greater than or less than a known value. It provides insight into the question of a direction-specific hypothesis, different from a two-tailed test which would check for any significant difference irrespective of direction.
For the Caribbean wedding cost exercise, the alternative hypothesis \( H_a: \mu < 10\) specifies a "less than" test, making it one-tailed. We're solely interested in whether costs are significantly lower than $10,000.
In practical terms, this affects how we interpret the t-statistic and the critical value from the t-distribution chart. A one-tailed test at a 0.05 level leads to the critical value being more extreme (further from zero) compared to a two-tailed test, suggesting that the evidence needed to reject the null must be stronger. Thus, using a one-tailed test helps refine our conclusions about specific directions in hypothesis testing.

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Most popular questions from this chapter

A sample of 64 observations is selected from a normal population. The sample mean is 215 , and the sample standard deviation is \(15 .\) Conduct the following test of hypothesis using the .03 significance level. $$\begin{array}{l}H_{0}: \mu \geq 220 \\\H_{1}: \mu<220\end{array}$$ For Exercises 5-8: (a) State the null hypothesis and the alternate hypothesis. (b) State the decision rule. (c) Compute the value of the test statistic. (d) What is your decision regarding \(H_{0} ?\) (e) What is the \(p\) -value? Interpret it.

The Gallup Organization in Princeton, New Jersey, is one of the best-known polling organizations in the United States. They often combine with USA Today or CNN to conduct polls of current interest. They also maintain a website at: http://www.gallup.com/. Consult this website to find the most recent polling results on Presidential approval ratings. You may need to click on Fast Facts. Test whether the majority (more than 50 percent) approve of the President's performance. If the article does not report the number of respondents included in the survey, assume that it is \(1,000,\) a number that is typically used.

A spark plug manufacturer claimed that its plugs have a mean life in excess of 22,100 miles. Assume the life of the spark plugs follows the normal distribution. A fleet owner purchased a large number of sets. A sample of 18 sets revealed that the mean life was 23,400 miles and the standard deviation was 1,500 miles. Is there enough evidence to substantiate the manufacturer's claim at the .05 significance level?

Research at the University of Toledo indicates that 50 percent of the students change their major area of study after their first year in a program. A random sample of 100 students in the College of Business revealed that 48 had changed their major area of study after their first year of the program. Has there been a significant decrease in the proportion of students who change their major after the first year in this program? Test at the .05 level of significance.

During recent seasons, Major League Baseball has been criticized for the length of the games. A report indicated that the average game lasts 3 hours and 30 minutes. A sample of 17 games revealed the following times to completion. (Note that the minutes have been changed to fractions of hours, so that a game that lasted 2 hours and 24 minutes is reported as 2.40 hours.) $$\begin{array}{|lllllllll|}\hline 2.98 & 2.40 & 2.70 & 2.25 & 3.23 & 3.17 & 2.93 & 3.18 & 2.80 \\\2.38 & 3.75 & 3.20 & 3.27 & 2.52 & 2.58 & 4.45 & 2.45 & \\\\\hline\end{array}$$ Can we conclude that the mean time for a game is less than 3.50 hours? Use the .05 significance level.

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