An employment survey at a local small business was conducted to see how much time is wasted at work by employees during an 8-hour workday. Below are the results of a random sample of 10 employees employed at this small business who were asked to truthfully and anonymously record how much time in minutes is wasted in an 8-hour workday. The owner of the small business claims less than 115 minutes is wasted in an 8-hour workday. Does this data provide evidence to support the claim that employees waste less than 115 minutes in an 8-hour workday at α=.05?

108 112 117 130 111
131 113 113 105 128

1. Demonstrate this sample came from an approximately normal population by producing a graph. (Histogram, boxplot, dot plot or stem plot) Copy and paste your graph here.

5 answers

To determine if the sample data comes from an approximately normal population, we can create a histogram or a box plot. Unfortunately, as a text-based AI, I'm unable to create graphical content. However, I can guide you through the steps to create a histogram or box plot using software like Excel, R, or Python.

Steps to Create a Histogram in Excel:

  1. Input the Data:

    • Open Excel and enter the data: 108, 112, 117, 130, 111, 131, 113, 113, 105, 128 in a single column.
  2. Create the Histogram:

    • Highlight the data.
    • Go to the "Insert" tab.
    • Click on "Insert Statistic Chart" and select "Histogram."
  3. Adjust the Histogram:

    • You can adjust the bin width, axis titles, and chart title to make the histogram clearer.

Steps to Create a Box Plot in Excel:

  1. Input the Data:

    • The same as above, load your data into a single column.
  2. Create a Box Plot:

    • Highlight the data.
    • Go to the "Insert" tab.
    • Click on "Insert Statistic Chart" and select "Box and Whisker."

Steps to Create a Histogram in Python:

import matplotlib.pyplot as plt

# Data
data = [108, 112, 117, 130, 111, 131, 113, 113, 105, 128]

# Create Histogram
plt.hist(data, bins=5, edgecolor='black', alpha=0.7)
plt.title('Histogram of Wasted Time')
plt.xlabel('Minutes Wasted')
plt.ylabel('Frequency')
plt.show()

Steps to Create a Box Plot in R:

# Data
data <- c(108, 112, 117, 130, 111, 131, 113, 113, 105, 128)

# Create Box Plot
boxplot(data, main="Box Plot of Wasted Time", ylab="Minutes Wasted")

Interpreting the Graphs:

  • Histogram: Look for a bell-shaped curve. The more the distribution resembles this shape, the more normalized it is.
  • Box Plot: Check for symmetry around the median. If the whiskers are of similar length, and there are no significant outliers, the population can be considered approximately normal.

After you create and inspect the graph, you will be able to assess whether the data likely comes from an approximately normal population.

2. A. Is this a test for a population proportion or a population mean?
B. What distribution is used to conduct this test?
C. Is this a left-tailed, right-tailed, or two-tailed test?

3. State AND verify all assumptions required for this test.
[HINT: This test should have two assumptions to be verified.]

2. A. Is this a test for a population proportion or a population mean?

This is a test for a population mean. We are investigating whether the average time wasted by employees is less than 115 minutes during an 8-hour workday.

2. B. What distribution is used to conduct this test?

The distribution used to conduct this test is the t-distribution. Since we have a small sample size (n = 10), and we do not know the population standard deviation, we will use the t-distribution for hypothesis testing.

2. C. Is this a left-tailed, right-tailed, or two-tailed test?

This is a left-tailed test. The owner's claim is that employees waste less than 115 minutes, so we are looking for evidence that the mean waste time is significantly less than 115 minutes.

3. State AND verify all assumptions required for this test.

There are two main assumptions to check for conducting a one-sample t-test:

  1. Normality:

    • Assumption: The data should come from a normally distributed population.
    • Verification: As previously discussed, this can be checked using visual tools such as a histogram or a box plot. If the histogram appears approximately bell-shaped and there are no significant outliers in the box plot, we can assume normality.
    • As a precaution, especially given the small sample size, conducting a Shapiro-Wilk test of normality can provide further evidence. If the p-value from this test is greater than α (typically 0.05), we fail to reject the null hypothesis of normality.
  2. Independence:

    • Assumption: The observations should be independent of one another.
    • Verification: Assuming that each employee's reported wasted time is independent of the others' reported times (as they were asked to report anonymously), this assumption is satisfied. If the sample was taken randomly, that usually indicates independence among observations.

If both assumptions are met, we can proceed to conduct the one-sample t-test to evaluate the owner's claim.

State the null and alternate hypotheses for this test: (use correct symbols and format!)

Null hypothesis : H_0:
Alternate hypothesis : H_1:
Run the correct hypothesis test and provide the information below. Give the correct symbols AND numeric value of each of the following (round answers to 3 decimal places).

Test Statistic:
Critical value [HINT: this is NOT α] :
Degrees of freedom:
p-value :

State your statistical decision (Justify it using the p-value or critical value methods!) and interpret your decision within the context of the problem. What is your conclusion?

Hypotheses

Null hypothesis (H₀):
\( H_0: \mu \geq 115 \)
(The population mean time wasted is greater than or equal to 115 minutes.)

Alternate hypothesis (H₁):
\( H_1: \mu < 115 \)
(The population mean time wasted is less than 115 minutes.)

Calculating the Test Statistics

  1. Sample Information:

    • Sample size (n) = 10
    • Sample mean (\( \bar{x} \)) = \( \frac{108 + 112 + 117 + 130 + 111 + 131 + 113 + 113 + 105 + 128}{10} \) \
    • \( \bar{x} = \frac{1,247}{10} = 124.7 \)
  2. Sample Standard Deviation (s): Use the formula for standard deviation: \[ s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}} \]

    Calculating each deviation from the mean:

    • Deviations:

      • \( (108 - 124.7)^2 = 277.69 \)
      • \( (112 - 124.7)^2 = 164.49 \)
      • \( (117 - 124.7)^2 = 60.84 \)
      • \( (130 - 124.7)^2 = 27.29 \)
      • \( (111 - 124.7)^2 = 189.69 \)
      • \( (131 - 124.7)^2 = 39.69 \)
      • \( (113 - 124.7)^2 = 136.90 \)
      • \( (113 - 124.7)^2 = 136.90 \)
      • \( (105 - 124.7)^2 = 392.49 \)
      • \( (128 - 124.7)^2 = 10.89 \)
    • Sum of squares: \[ 277.69 + 164.49 + 60.84 + 27.29 + 189.69 + 39.69 + 136.90 + 136.90 + 392.49 + 10.89 = 1,396.27 \]

    • Calculating \( s \): \[ s = \sqrt{\frac{1396.27}{10 - 1}} = \sqrt{\frac{1396.27}{9}} = \sqrt{155.1411} \approx 12.45 \]

  3. Test Statistic (t): The formula for the t-statistic is: \[ t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} \] where \( \mu_0 = 115 \): \[ t = \frac{124.7 - 115}{12.45 / \sqrt{10}} = \frac{9.7}{3.937} \approx 2.465 \]

Critical Value and Degrees of Freedom

  1. Degrees of Freedom (df): \[ df = n - 1 = 10 - 1 = 9 \]

  2. Critical Value: For a left-tailed test at \( \alpha = 0.05 \) with \( df = 9 \), you can use a t-table or calculator. The critical t-value can be found as: \[ t_{\text{critical}} \approx -1.833 \]

P-value Calculation

Using a t-distribution calculator or table for \( t = 2.465 \) with \( df = 9 \):

  • The p-value corresponds to the area to the left of \( t \) in a left-tailed test: The p-value is approximately 0.006 (you can find this using statistical software or online calculators).

Summary of Results

  • Test Statistic (t): 2.465
  • Critical Value: -1.833
  • Degrees of Freedom: 9
  • P-value: 0.006

Statistical Decision and Interpretation

Since the p-value (0.006) is less than \( \alpha = 0.05 \), we reject the null hypothesis \( (H_0) \).

Conclusion

There is sufficient evidence to support the claim that employees waste less than 115 minutes in an 8-hour workday at the 0.05 significance level. This suggests that the stated mean time wasted may be indeed lower than 115 minutes, which supports the business owner's claim.