• Objective

To estimate the average customer satisfaction with the Professional job category (μ) based on sample data, with a 95% confidence level.

• Sample Data & Statistics

Number of respondents (n) = 120 Mean satisfaction (x̄) = 7.85 Sample standard deviation (s) = 1.25

• Confidence Level

Confidence Level = 95%

Significance level (α) = 0.05

• Method Used

The t-distribution is used because the population standard deviation is unknown. The general formula is:

\[ CI = \bar{x} \pm t_{\alpha/2, df} \cdot \left( \frac{s}{\sqrt{n}} \right) \] Degrees of freedom (df) = n - 1 = 119

• Calculation

Critical t-value: t_{0.025,119} ≈ 1.98

Standard Error (SE):

\[ SE = \frac{s}{\sqrt{n}} = \frac{1.25}{\sqrt{120}} \approx 0.114 \]

Margin of Error (ME):

\[ ME = t \cdot SE = 1.98 \times 0.114 \approx 0.23 \] Confidence Interval:

\[ CI = \bar{x} \pm ME = 7.85 \pm 0.23 \Rightarrow [7.62,\ 8.08] \]

• Interpretation

The average customer satisfaction score for Professionals ranges from 7.62 to 8.08 with a 95% confidence level. This means that if this study were repeated multiple times, approximately 95% of the resulting interval would encompass the true average satisfaction score for the Professional population.

• Additional Notes

The general population mean (7.62) falls right at the lower end of the interval.

This indicates that Professional customers tend to have higher satisfaction, although the difference is not significant.

• Hypothesis:

\[H_0: \mu_{\text{Prof}} = \mu_{\text{Pop}} \quad \text{vs} \quad H_1: \mu_{\text{Prof}} > \mu_{\text{Pop}}\] one-way test because the focus is “higher”.

Data and Assumptions

• Sample Statistics (Professional \[n = 120, \bar{x} = 7.85, s = 1.25.\]

• Population mean (of the entire sample) \(\mu_{\text{Pop}} = 7.62.\)

• Key assumptions:

  • Independence among respondents.

  • Random sample of Professional customers.

  • The t-approximation is valid because n is large enough; robust to minor deviations from normality.

• Test Selection and Test Statistics

Test Used: One-Sample t-test against the reference value \(\mu_{\text{Pop}} = 7.62\).

Test Statistics:

\[t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} = \frac{7.85 - 7.62}{1.25/\sqrt{120}}\]

\[s/\sqrt{n} = \frac{1.25}{\sqrt{120}} \approx 0.114, \quad t \approx \frac{0.23}{0.114} \approx 2.02\]

Degrees of freedom:\(\text{df} = n - 1 = 119\)

Test decision (α = 0.05)

• One-tailed test: compare t with \(t_{0.95, 119} \approx 1.66\)

• Since \(2.02 > 1.66, \text{ Reject } H_0\)

• p-value (one-tailed): \(\approx 0.022 \to < 0.05\)

• \(( \text{If two-tailed: } p \approx 0.045, \text{ remains significant at } 5\%).\)

Estimated effects and practical significance

• Cohen’s d:

\[d = \frac{\bar{x} - \mu_0}{s} = \frac{0.23}{1.25} \approx 0.18\]

ESmall effect—statistically significant, but the increase in satisfaction is moderate in practical terms.

• Confidence interval for the difference in means (two-tailed, 95%):

\[( \bar{x} - \mu_0 ) \pm t_{0.975, 119} \cdot \frac{s}{\sqrt{n}} = 0.23 \pm 1.98 \cdot 0.114 \\\]

\[\approx 0.23 \pm 0.23 \implies [0.00, 0.46]\] The lower bound is close to zero—consistent with a small effect.

Robustness check (brief)

• Mild normality: with n = 120, the t-test is quite robust.

• Outliers: check the boxplot/histogram of Professional satisfaction scores; if there are strong outliers, consider a nonparametric test (discussed separately).

• Equality of variance is not relevant for the one-sample t-test, but is important when moving on to comparing two independent groups.

• Inferential Conclusion

With a one-tailed test of alpha = 0.05, the average satisfaction of Professional customers is significantly higher than the population average (t = approx 2.02, p = approx 0.022). While significant, the effect size is small (d = approx 0.18), so practical implications need to be considered in conjunction with the business context (e.g., service segmentation or loyalty programs targeting Professionals).

• Interpretation

Mean customer satisfaction among Professionals is significantly higher than the population average. However, the effect size is relatively small (Cohen’s d ≈ 0.18), so while statistically significant, this increase in satisfaction is not significant in practical terms.

• Method Used

The method used was the Wilcoxon Signed-Rank Test, a nonparametric test used to compare the median of a sample with a specific reference value. This test was chosen because customer satisfaction data is not assumed to be normally distributed.

• Hypotesis

H_0:\(\ \tilde{X}{\text{Professional}} \le \tilde{X}{\text{Population}}\)

H_1:\(\ \tilde{X}{\text{Professional}} > \tilde{X}{\text{Population}}\)

\(\alpha = 0{,}05\)

• Calculation Steps

The difference is calculated using

\[d_i = X_i - \tilde{X}_0\]

Observations with \(d_i = 0\) are excluded from the Wilcoxon Signed-Rank analysis, and then the absolute values of \(|d_i|\) are ranked. The sum of the positive signed ranks is used as the Wilcoxon test statistic.

• Calculation Results

\[\tilde{X}_{\text{Population}} = 3{,}7383\]

\[\tilde{X}_{\text{Professional}} = 4{,}5465\]

\[W = 7\]

\[p\text{-value} = 0{,}3125\]

• Test Decision

\[p\text{-value} > \alpha \Rightarrow H_0 \text{fail to reject}\]

• Conclusion

Based on the Wilcoxon Signed-Rank test, the statistical value obtained was \(W = 7\) with \(p\text{-value} = 0{.}3125\). Since \(p\text{-value} > \alpha = 0{.}05\), there is insufficient statistical evidence to state that customers in the Professional job category have a higher level of satisfaction than the overall customer population.