Problem 39 - Basic Statistics for Business and Economics with Student

Question

eGolf.com receives an average of 6.5 returns per day from online shoppers. For a sample of 12 days, it received the following number of returns $$\begin{array}{|llllllllllll|}\hline 0 & 4 & 3 & 4 & 9 & 4 & 5 & 9 & 1 & 6 & 7 & 10 \\\\\hline\end{array}$$ At the .01 significance level, can we conclude the mean number of returns is less than \(6.5 ?\)

Step-by-Step Solution

Answer

The mean number of returns is not significantly less than 6.5 at the 0.01 level.

Step 1: Define null and alternative hypotheses

The null hypothesis (H_0) is that the true mean number of returns is equal to 6.5, while the alternative hypothesis (H_1) is that the mean number of returns is less than 6.5.\[ H_0: \mu = 6.5 \]\[ H_1: \mu < 6.5 \]

Step 2: Calculate sample mean

The sample mean is calculated by summing all the sample returns and dividing by the number of days. The sample returns are 0, 4, 3, 4, 9, 4, 5, 9, 1, 6, 7, 10.\[ \bar{x} = \frac{0 + 4 + 3 + 4 + 9 + 4 + 5 + 9 + 1 + 6 + 7 + 10}{12} = \frac{62}{12} = 5.17 \]

Step 3: Compute sample standard deviation

Find the standard deviation (S) of the sample data to measure the dispersion.\[ S = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}} \]Where (x_i) is each data point, (\bar{x}) is the sample mean, 5.17, and (n) is the number of days, 12.Perform the calculation:\[ S = \sqrt{\frac{(0-5.17)^2 + (4-5.17)^2 + \ldots + (10-5.17)^2}{11}} \approx 2.88 \]

Step 4: Calculate the test statistic

Use the formula for the t-test statistic:\[ t = \frac{\bar{x} - \mu}{S/\sqrt{n}} \]Substitute known values (\bar{x} = 5.17, \mu = 6.5, S = 2.88, and (n = 12)) into the formula:\[ t = \frac{5.17 - 6.5}{2.88/\sqrt{12}} = \frac{-1.33}{0.832} \approx -1.60 \]

Step 5: Determine the critical value

Using the t-distribution table, find the critical t-value at the 0.01 significance level with 11 degrees of freedom (n-1). For a one-tailed test, the critical value is -2.718.

Step 6: Compare the test statistic to the critical value

Compare the calculated t-value with the critical t-value:\[ t = -1.60, \quad t_{critical} = -2.718 \]Since t > t_{critical}, we fail to reject the null hypothesis.

Step 7: Conclusion

At the 0.01 significance level, the evidence is not sufficient to conclude that the mean number of returns is less than 6.5.

Key Concepts

Sample Mean

The sample mean is a fundamental statistical measure used to estimate the average of a set of values. It's calculated by adding up all the values and dividing by the number of values in the set, known as the sample size.

In the context of eGolf.com's returns, the sample mean helps us understand the average number of returns per day over a given period. The importance of calculating the sample mean lies in its ability to give us a snapshot of sample data behavior, such as return frequency.

For instance, by summing up all the return counts from the 12 days and dividing by 12, we find the sample mean for the number of returns is 5.17. This means, on average, eGolf.com received approximately 5.17 returns per day during this period, which we use as a basis for further analysis.

Standard Deviation

Standard deviation measures how much individual data points diverge from the sample mean. It's a way to quantify the variability or dispersion in a sample dataset.

To compute the standard deviation, we first determine the difference between each data point and the sample mean, square those differences, sum them up, and divide by the number of data points minus one.

Lastly, we take the square root. This result gives an approximation of the dataset's internal scatter.

For the number of returns to eGolf.com, the standard deviation was calculated to be approximately 2.88. A higher standard deviation would indicate a wide variety of daily returns, while a lower value would suggest returns are more consistent from day to day.

T-Test

A t-test is a statistical procedure used to determine if there is a significant difference between a sample mean and a known value or another sample mean. It helps us assess whether observed data deviates from expectations due to random chance.

When examining eGolf.com's returns, the t-test evaluates whether the observed average return count (sample mean) of 5.17 is significantly lower than the claimed average of 6.5.

The test statistic is calculated using the formula: \[ t = \frac{\bar{x} - \mu}{S/\sqrt{n}} \]where - \( \bar{x} \) is the sample mean- \( \mu \) is the claimed mean- \( S \) is the standard deviation- \( n \) is sample size.

For eGolf.com's data, the calculated t-value was approximately -1.60, indicating the difference between the sample mean and 6.5, but not by enough considering the sample's variability.

Significance Level

The significance level is a threshold we set to determine the strength of evidence required to reject the null hypothesis in a statistical test. Common significance levels are 0.05, 0.01, and 0.10.

The lower the level, the stricter the criteria to reject the null hypothesis. For eGolf.com, we use a 0.01 significance level, meaning we accept only a 1% chance of incorrectly rejecting the null hypothesis.

In hypothesis testing, this level helps to define the rejection region. Using the t-distribution, we find the critical value that corresponds to this level, and compare it against our test statistic.

For eGolf.com's example, the critical t-value at 0.01 level with 11 degrees of freedom (sample size minus one) was -2.718. Since our calculated t-value was -1.60, which is greater than -2.718, we fail to reject the null hypothesis, reinforcing that there's insufficient evidence to claim that returns average less than 6.5 per day.