Problem 54 - Basic Statistics for Business and Economics with Student

Question

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 thousands of dollars for a sample of 8 Caribbean weddings $$\begin{array}{|llllllll|}\hline 9.7 & 9.4 & 11.7 & 9.0 & 9.1 & 10.5 & 9.1 & 9.8 \\\\\hline\end{array}$$ At the .05 significance level is it reasonable to conclude the mean wedding cost is less than \(\$ 10,000\) as advertised?

Step-by-Step Solution

Answer

At the 0.05 significance level, there is not enough evidence to conclude the mean wedding cost is less than $10,000.

Step 1: State the Hypotheses

First, define the null and alternative hypotheses. The null hypothesis, \(H_0\), states that the mean cost of a Caribbean wedding is \( \mu = 10,000 \) dollars. The alternative hypothesis, \(H_a\), is that the mean cost is less than \(10,000\) dollars, \( \mu < 10,000 \).

Step 2: Calculate the Sample Mean

Convert the wedding costs into thousands and calculate the sample mean, \( \bar{x} \). The costs are already in thousands of dollars: \(9.7, 9.4, 11.7, 9.0, 9.1, 10.5, 9.1, 9.8\). \[\bar{x} = \frac{9.7 + 9.4 + 11.7 + 9.0 + 9.1 + 10.5 + 9.1 + 9.8}{8} = 9.663data{.}\]

Step 3: Calculate the Sample Standard Deviation

Use the formula for sample standard deviation. First, calculate the deviations from the mean (\(\bar{x}\)) for each sample \(x_i\), square these deviations, sum them, and divide by \(n-1\), where \(n\) is the number of samples.\[s = \sqrt{\frac{(9.7-9.663)^2 + (9.4-9.663)^2 + (11.7-9.663)^2 + \cdots + (9.8-9.663)^2}{7}}\approx 0.842data{.}\]

Step 4: Calculate the Test Statistic

Calculate the test statistic using the formula for the t-score:\[t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}}\]Where \(\mu_0 = 10\), \(\bar{x} = 9.663\), \(s \approx 0.842\), and \(n = 8\):\[t = \frac{9.663 - 10}{0.842 / \sqrt{8}} \approx -1.265data{.}\]

Step 5: Determine the Critical Value and Decision

Use a t-distribution table or calculator to find the t-critical value for \(n - 1 = 7\) degrees of freedom at the 0.05 significance level for a one-tailed test, which is approximately \(-1.895\). Compare the test statistic to the critical value. Since \(-1.265 > -1.895\), fail to reject the null hypothesis.

Key Concepts

Significance Level

The significance level is an important concept in hypothesis testing, helping us determine the threshold for rejecting a null hypothesis. In simpler terms, it is the probability of rejecting the null hypothesis when it is actually true. This is also called the 'Type I error rate'.

For example, in the wedding cost problem, the significance level chosen is 0.05 or 5%. This means we are willing to accept a 5% chance of incorrectly concluding that the average wedding cost is less than $10,000 when it is not.

• A smaller significance level (like 0.01) means being more stringent about rejecting the null hypothesis, ensuring higher confidence in our decision.
• Conversely, a larger significance level (like 0.1) increases the chance of a Type I error, making it easier to reject the null hypothesis.

Choosing the right significance level involves balancing the risks of errors with the need for statistical sensitivity.

Sample Mean

The sample mean is a crucial statistical measure that helps summarize the central tendency of a data sample. It acts as an estimate of the true population mean, providing insights with limited data.

For instance, in the Caribbean weddings scenario, we calculate the sample mean from the eight listed wedding costs. By adding all the costs together and dividing by the number of data points (8 in this case), we find the average cost.

• The formula used here is: \[\bar{x} = \frac{\text{sum of all costs}}{\text{number of weddings}}.\]
• Our calculated mean of approximately 9.663 provides a snapshot of the data, suggesting that the average wedding cost in the sample is under the $10,000 mark.

The sample mean is important because it allows us to test hypotheses based on the observed data, aiming to make broader predictions about an entire population.

Standard Deviation

Standard deviation measures the spread or dispersion of a set of data points. It tells us how much the values in our sample differ from the sample mean.

In the exercise, we calculated the standard deviation to check the variability of wedding costs. A lower standard deviation would mean that the costs are close to the sample mean, and a higher one would indicate more variability.

• To find the standard deviation, calculate the difference of each data point from the mean, square those differences, sum them up, and then take the square root of the result divided by the number of data points minus one.
• The formula is:\[s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}.\]
• In our problem, we got approximately 0.842, showing a moderate level of variability in wedding costs.

Understanding standard deviation helps us grasp how typical or atypical individual data points are compared to the average, shaping our hypothesis testing decisions.

T-Test

The t-test is a statistical method used to compare the sample mean against a known value or another sample mean. It helps determine if there are significant differences between the groups being compared.

In our problem, the t-test assesses whether the sample mean wedding cost significantly differs from the advertised $10,000. Utilizing the sample mean, standard deviation, and sample size, we calculate the t-test statistic.

• The formula used: \[t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}}\]where \(\mu_0\) is the hypothesized population mean.
• The calculated t-value is compared to a t-distribution table to determine whether it falls in the critical region.
• For our case, the t-value of approximately -1.265 did not exceed the critical value of -1.895 at the 0.05 significance level, meaning we fail to reject the null hypothesis.

The t-test is essential in deciding claims about population means when the sample size is small or when population variance is unknown. It forms the backbone of many hypothesis tests in real-world scenarios.