Problem 11
Question
The Rocky Mountain district sales manager of Rath Publishing, Inc., a college textbook publishing company, claims that the sales representatives make an average of 40 sales calls per week. Several reps say that this estimate is too low. To investigate, a random sample of 28 sales representatives reveals that the mean number of calls made last week was \(42 .\) The standard deviation of the sample is 2.1 calls. Using the .05 significance level, can we conclude that the mean number of calls per salesperson per week is more than \(40 ?\)
Step-by-Step Solution
Verified Answer
Yes, the mean number of calls is significantly more than 40.
1Step 1: Identify the Hypotheses
We need to establish our null hypothesis \( H_0 \) and alternative hypothesis \( H_1 \). Here, the null hypothesis is that the true mean number of calls is 40, \( H_0: \mu = 40 \). The alternative hypothesis is that the mean number of calls is greater than 40, \( H_1: \mu > 40 \).
2Step 2: Determine the Significance Level and Distribution
The significance level (\( \alpha \)) is 0.05. We will use a one-sample t-test because we have one sample with a small size (less than 30) and the population standard deviation is unknown.
3Step 3: Calculate the Test Statistic
The test statistic for a t-test is calculated as \[ t = \frac{\bar{x} - \mu}{s/\sqrt{n}} \]where \( \bar{x} = 42 \) is the sample mean, \( \mu = 40 \) is the population mean under the null hypothesis, \( s = 2.1 \) is the sample standard deviation, and \( n = 28 \) is the sample size.Substituting these values in, \[ t = \frac{42 - 40}{2.1/\sqrt{28}} \].
4Step 4: Compute the t-value
Continuing from the formula, \[ t = \frac{2}{2.1/\sqrt{28}} = \frac{2}{2.1/5.2915} \approx \frac{2}{0.397} \approx 5.04 \].
5Step 5: Determine the Critical t-value
For a one-tailed t-test with 27 degrees of freedom (\( n - 1 = 28 - 1 = 27 \)), and \( \alpha = 0.05 \), we find the critical t-value from t-distribution tables. The critical t-value is approximately 1.703.
6Step 6: Compare the Test Statistic to the Critical Value
Since the calculated t-value of 5.04 is greater than the critical value of 1.703, we reject the null hypothesis. This means there is sufficient evidence to conclude that the mean number of sales calls per week is greater than 40.
Key Concepts
One-sample t-testSignificance LevelCritical t-valueTest Statistic
One-sample t-test
The one-sample t-test is a statistical method used to determine whether there is a significant difference between the mean of a sample and a known or hypothesized population mean. In this exercise, we want to check if sales representatives on average make more than 40 calls per week. The null hypothesis states that the sample mean equals the population mean of 40. Meanwhile, the alternative hypothesis suggests that the sample mean is greater than 40. The tool of a one-sample t-test is well-suited for this investigation. We use it because we have a small sample size of 28, which is less than 30, and we do not know the population standard deviation but have the sample standard deviation instead.
In essence, this test evaluates whether any observed difference between the sample and population mean is statistically significant or could have just occurred by chance. This allows decision-making with quantified confidence levels.
In essence, this test evaluates whether any observed difference between the sample and population mean is statistically significant or could have just occurred by chance. This allows decision-making with quantified confidence levels.
Significance Level
The significance level is a threshold that helps us determine whether our result is statistically significant. In this exercise, it is set at 0.05, or 5%. This means we are allowing a 5% chance of incorrectly rejecting the null hypothesis, which is a common practice in hypothesis testing. By setting a significance level, we're defining the acceptable probability of committing a "Type I error," where we might erroneously conclude that there is an effect or difference when there isn't one.
Choosing a significance level relies on how much risk you are willing to take. Typically, a lower significance level protects against drawing incorrect conclusions—the lower the alpha, the less likely you are to have a false positive. In practice, a 0.05 level is used in many fields as a standard convention, providing a balanced approach between being too conservative and too liberal when evaluating results.
Choosing a significance level relies on how much risk you are willing to take. Typically, a lower significance level protects against drawing incorrect conclusions—the lower the alpha, the less likely you are to have a false positive. In practice, a 0.05 level is used in many fields as a standard convention, providing a balanced approach between being too conservative and too liberal when evaluating results.
Critical t-value
The critical t-value is a point on the t-distribution that is compared against the test statistic to decide whether to reject the null hypothesis. In our case, we calculate this value considering the significance level and the degrees of freedom (df). For a one-tailed test with a significance level of 0.05 and 27 degrees of freedom (since df = sample size - 1), the critical t-value from standard t-distribution tables is about 1.703.
This value serves as a benchmark in hypothesis testing. If the calculated test statistic is greater than the critical t-value, it indicates that the result is unlikely under the null hypothesis—which leads to its rejection. The critical t-value helps visually represent the boundary of the rejection region in a t-distribution curve, making it clearer to decide if our sample provides enough evidence against the null hypothesis.
This value serves as a benchmark in hypothesis testing. If the calculated test statistic is greater than the critical t-value, it indicates that the result is unlikely under the null hypothesis—which leads to its rejection. The critical t-value helps visually represent the boundary of the rejection region in a t-distribution curve, making it clearer to decide if our sample provides enough evidence against the null hypothesis.
Test Statistic
The test statistic is a calculated value that determines whether we should reject the null hypothesis. It is a standardized value derived from how much our sample data deviates from the null hypothesis expectation. In this exercise, we compute the test statistic using the formula:
\[ t = \frac{\bar{x} - \mu}{s/\sqrt{n}} \]
Here, \( \bar{x} \) is the sample mean (42), \( \mu \) is the population mean under the null hypothesis (40), \( s \) is the sample standard deviation (2.1), and \( n \) is the sample size (28). By plugging in these numbers, we obtain a test statistic of approximately 5.04.
This value is essential when comparing against the critical t-value. The higher this test statistic, the more likely that the observed data differs significantly from the null hypothesis assumption. In simpler terms, for our calculated t of 5.04 being greater than the critical t-value of 1.703, we have grounds to reject the null hypothesis and conclude with confidence that sales reps typically make more than 40 calls weekly.
\[ t = \frac{\bar{x} - \mu}{s/\sqrt{n}} \]
Here, \( \bar{x} \) is the sample mean (42), \( \mu \) is the population mean under the null hypothesis (40), \( s \) is the sample standard deviation (2.1), and \( n \) is the sample size (28). By plugging in these numbers, we obtain a test statistic of approximately 5.04.
This value is essential when comparing against the critical t-value. The higher this test statistic, the more likely that the observed data differs significantly from the null hypothesis assumption. In simpler terms, for our calculated t of 5.04 being greater than the critical t-value of 1.703, we have grounds to reject the null hypothesis and conclude with confidence that sales reps typically make more than 40 calls weekly.
Other exercises in this chapter
Problem 9
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