Problem 20 - Basic Statistics for Business and Economics with Student

Question

Hugger Polls contends that an agent conducts a mean of 53 in-depth home surveys every week. A streamlined survey form has been introduced, and Hugger wants to evaluate its effectiveness. The number of in-depth surveys conducted during a week by a random sample of agents are: $$\begin{array}{|lllllllllllllll|}\hline 53 & 57 & 50 & 55 & 58 & 54 & 60 & 52 & 59 & 62 & 60 & 60 & 51 & 59 & 56 \\\\\hline\end{array}$$ At the .05 level of significance, can we conclude that the mean number of interviews conducted by the agents is more than 53 per week? Estimate the \(p\) -value.

Step-by-Step Solution

Answer

The mean number of interviews is significantly more than 53 per week.

Step 1: Define Null and Alternative Hypotheses

The null hypothesis (H0) is that the mean number of interviews is 53: \( H_0: \mu = 53 \). The alternative hypothesis (H1) is that the mean number of interviews is greater than 53: \( H_1: \mu > 53 \).

Step 2: Calculate Sample Mean and Standard Deviation

Calculate the sample mean (\( \bar{x} \)) and sample standard deviation (s) for the data. The data is \([53, 57, 50, 55, 58, 54, 60, 52, 59, 62, 60, 60, 51, 59, 56]\). \( \bar{x} = \frac{53 + 57 + 50 + 55 + 58 + 54 + 60 + 52 + 59 + 62 + 60 + 60 + 51 + 59 + 56}{15} = \frac{896}{15} \approx 59.73 \).

Step 3: Conduct a One-Sample t-Test

With the sample mean (\( \bar{x} = 59.73 \)), sample size (n = 15), sample standard deviation (\(s\)), and population mean (\(\mu = 53\)), perform a one-sample t-test. The t-statistic is calculated using \( t = \frac{\bar{x} - \mu}{s / \sqrt{n}} \).

Step 4: Determine Critical t-Value and p-Value

At the 0.05 level of significance for a one-tailed test with \( n - 1 = 14 \) degrees of freedom, find the critical t-value from a t-distribution table. Compute the p-value associated with the calculated t-statistic.

Step 5: Compare t-Statistic and Critical t-Value

Compare the calculated t-statistic to the critical t-value. If the t-statistic is greater, reject the null hypothesis. Also, compare the p-value to 0.05.

Step 6: Draw Conclusion

Based on the comparison, conclude whether there is sufficient evidence to reject the null hypothesis and claim that the mean number of interviews is more than 53 per week.

Key Concepts

Understanding the t-test

The t-test is a statistical method used to determine whether there is a significant difference between the means of two groups. In our case, it's a one-sample t-test to see if the mean number of surveys conducted by agents with the new form is more than the claimed 53 per week. This test is essential because it helps us make data-driven decisions based on sample data rather than assumptions.

• The difference between the sample mean and the known population mean is analyzed.
• You use the t-test to see if this difference is statistically significant.
• The test calculates a t-statistic, which can then be compared to a critical value, or used to find a p-value.

This test is especially useful when dealing with smaller sample sizes, where the normal distribution is not a good fit.

Significance Level

The significance level, often denoted as alpha (\( \alpha \)), is a threshold that the p-value must pass to reject the null hypothesis. In our exercise, the significance level is set at 0.05. This means there is a 5% risk of concluding that a difference exists when there really is none.

• It's a way of balancing the risk of a Type I error (false positive).
• In hypothesis testing, if the p-value is less than or equal to this significance level, we reject the null hypothesis.
• A lower significance level means stricter evidence is needed to reject the null hypothesis.

Choosing the significance level is a critical step in hypothesis testing as it directly impacts the strength and reliability of your conclusion.

Role of the Sample Mean

In this context, the sample mean helps us gauge the average performance based on the sample data. It's a statistic that summarizes the data collected from agents. Our calculated sample mean of \( 59.73 \) sheds light on how agents are performing on average with the new form.

• The sample mean (\( \bar{x} \)) is calculated by adding all the data points together and dividing by the number of points.
• It provides a central value for the data, offering insights into the overall trend.
• We compare the sample mean to 53 (claimed mean) to assess if there is a significant difference.

A higher sample mean than the claimed population mean suggests that the new form may be effective.

Decoding the p-value

The p-value in hypothesis testing helps determine the probability of observing the results, or something more extreme, assuming the null hypothesis is true. If our p-value is less than or equal to the significance level, we have grounds to reject the null hypothesis. It's like a litmus test for statistical significance.

• A small p-value (< 0.05 in our context) indicates that the observed data is unlikely under the null hypothesis.
• It helps researchers decide whether to reject or fail to reject the null hypothesis.
• The p-value is derived from the t-statistic and gives a quantitative measure of the evidence against the null hypothesis.

By calculating the p-value, we can efficiently judge the effectiveness of the new survey form based on this sample.