Consider the e-billing case. The mean and the standard deviation of the sample of n = 65 payment times are  x⎯⎯
  = 18.7934 and s = 3.9255. Test H0: μ = 19.6 versus Ha: μ < 19.6 by setting α equal to .01 and using a critical value rule and assume normality of the population. (Round your "t" and "t0.01" answers to 3 decimal places and p-value answer to 4 decimal places. Negative values should be indicated by a minus sign. Use a statistical software package - e.g., Minitab, MegaStat, etc., to derive the p-value.)
 



 



To test the hypothesis \( H_0: \mu = 19.6 \) versus \( H_a: \mu < 19.6 \) using the given data, we'll follow these steps:

1. **State the hypotheses:**
   - Null hypothesis (\( H_0 \)): \( \mu = 19.6 \)
   - Alternative hypothesis (\( H_a \)): \( \mu < 19.6 \)

2. **Determine the test statistic:**
   The test statistic for a one-sample t-test is given by:
   \[
   t = \frac{\bar{x} - \mu}{s / \sqrt{n}}
   \]
   where:
   - \(\bar{x} = 18.7934\)
   - \(\mu = 19.6\)
   - \(s = 3.9255\)
   - \(n = 65\)

   Plugging in these values:
   \[
   t = \frac{18.7934 - 19.6}{3.9255 / \sqrt{65}}
   \]

3. **Calculate the test statistic:**
   \[
   t = \frac{18.7934 - 19.6}{3.9255 / \sqrt{65}} = \frac{-0.8066}{0.4864} \approx -1.658
   \]

4. **Find the critical value:**
   For a one-tailed test at \(\alpha = 0.01\) with \(df = n - 1 = 64\), we use the t-distribution table to find the critical value \( t_{0.01} \).
   
   Using a t-table or software:
   \[
   t_{0.01, 64} \approx -2.387
   \]

5. **Determine the p-value:**
   Using a t-distribution table or statistical software, we find the p-value for \( t = -1.658 \) with \(df = 64\).

   Using statistical software, the p-value is approximately:
   \[
   \text{p-value} \approx 0.0512
   \]

6. **Conclusion:**
   - Compare the test statistic to the critical value:
     \[
     t = -1.658 \quad \text{and} \quad t_{0.01} = -2.387
     \]
     Since \( t \) is not less than \( t_{0.01} \), we do not reject the null hypothesis \( H_0 \).
   - Compare the p-value to \(\alpha\):
     \[
     \text{p-value} = 0.0512 \quad \text{and} \quad \alpha = 0.01
     \]
     Since the p-value is greater than \(\alpha\), we do not reject the null hypothesis \( H_0 \).

**Conclusion:** At the \( \alpha = 0.01 \) significance level, we do not have sufficient evidence to reject the null hypothesis. Therefore, we do not conclude that the mean payment time is less than 19.6.

If you have any questions or want more details, feel free to ask.

Here are five more questions you might want to explore:
1. What is the power of the test given the effect size?
2. How would the results change if the sample size were larger?
3. What are the assumptions behind using the t-test in this context?
4. How do you calculate confidence intervals for the mean payment time?
5. Can we perform a similar hypothesis test using non-parametric methods?

**Tip:** Always ensure the normality assumption is reasonable before applying a t-test, especially with smaller sample sizes.

Explore how to conduct a one-sample t-test to compare the mean payment times against a given null hypothesis. Understand the significance level, critical values, and p-value calculation using statistical software.

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