Statistical Inference for Employee Satisfaction Index (ESI)

Q6. An international company is aware that the mean Employee Satisfaction Index (ESI) within its Denmark-based branches is excellent, being 90, on a scale from 0 to 100. A union representative wishes to ascertain if the same mean of 90 is achieved or not within the company’s U.S.-based branches. Here are some summary statistics and a histogram of the ESIs obtained from a random sample of 25 U.S. employees.

Background

Topic: Hypothesis Testing for Means (One-Sample t-Test)

This question tests your understanding of hypothesis testing for a population mean, including setting up hypotheses, checking assumptions, calculating test statistics and p-values, and interpreting confidence intervals. The histogram helps assess normality, which is important for the t-test.

Key Terms and Formulas

• Null Hypothesis ($H_0$): The mean ESI in U.S. branches is 90 ($\mu = 90$).

• Alternative Hypothesis ($H_A$): The mean ESI in U.S. branches is not 90 ($\mu \neq 90$).

• Test Statistic: $t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}$

• Confidence Interval: $\bar{x} \pm t^* \frac{s}{\sqrt{n}}$

• Assumptions: Random sample, approximately normal distribution (check histogram), independence.

Step-by-Step Guidance

1. Write the null and alternative hypotheses: $H_0: \mu = 90$ $H_A: \mu \neq 90$

2. Check assumptions: Is the sample random? Is the sample size large enough or is the data approximately normal? Use the histogram to assess normality.

3. Calculate the test statistic using the formula: $t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}$ Where $\bar{x} = 81.44$, $\mu_0 = 90$, $s = 6.056$, $n = 25$.

4. Find the p-value associated with your calculated t-statistic using the t-distribution with $n-1$ degrees of freedom.

5. Construct a 95% confidence interval for the mean ESI using: $\bar{x} \pm t^* \frac{s}{\sqrt{n}}$ Where $t^*$ is the critical value for 95% confidence and 24 degrees of freedom.

Try solving on your own before revealing the answer!

Final Answer:

1. Hypotheses: $H_0: \mu = 90$, $H_A: \mu \neq 90$ 2. Assumptions: The histogram shows approximate normality, sample is random. 3. Test statistic: $t \approx -7.07$ 4. p-value: $\approx 0.00000013$ (very small) 5. Conclusion: There is strong evidence that the mean ESI in U.S. branches is less than 90. 6. 95% Confidence Interval: $81.44 \pm 2.064 \times \frac{6.056}{\sqrt{25}} \approx [79.01, 83.87]$ 7. Interpretation: The interval does not include 90, so the union representative should advocate for better working conditions in the U.S. branches.