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Likert Scale
Likert Scale Analysis — Descriptive Stats, Cronbach's Alpha & Interpretation
Complete worked example: 5-item Likert scale (1–5) for 8 employees. Calculates item means, total score, Cronbach's alpha (α = .90), and interprets the scale level with APA-7 reporting.
📊 α = .90 (ممتاز) | M = 3.63/5 (مرتفع)
Step-by-step solution
📋 23 · Likert Scale Analysis — Descriptive Stats & Cronbach's Alpha
5-Item Job Satisfaction Scale — Full Analysis
Likert Scale
Research Question:
A researcher developed a 5-item job satisfaction scale (Likert 1–5) for 8 employees. Calculate descriptive statistics, Cronbach's alpha, and interpret the results.
| Employee | Item 1 | Item 2 | Item 3 | Item 4 | Item 5 | Total |
|---|---|---|---|---|---|---|
| 1 | 4 | 3 | 4 | 5 | 4 | 20 |
| 2 | 2 | 2 | 3 | 2 | 3 | 12 |
| 3 | 5 | 4 | 5 | 4 | 5 | 23 |
| 4 | 3 | 3 | 3 | 4 | 3 | 16 |
| 5 | 4 | 5 | 4 | 5 | 4 | 22 |
| 6 | 2 | 3 | 2 | 3 | 2 | 12 |
| 7 | 5 | 4 | 5 | 5 | 4 | 23 |
| 8 | 3 | 4 | 3 | 3 | 4 | 17 |
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1Total score descriptives:$$\bar{x}_{total} = \frac{20+12+23+16+22+12+23+17}{8} = \frac{145}{8} = 18.13, \quad SD = 4.36$$
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2Item mean (per-item average, scale 1–5):$$\bar{x}_{item} = \frac{18.13}{5} = 3.63 \quad \text{out of 5}$$
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3Cronbach's Alpha (internal consistency):$$\alpha = \frac{k}{k-1}\left(1 - \frac{\sum s_i^2}{s_T^2}\right) = \frac{5}{4}\left(1 - \frac{5.33}{19.0}\right) = \frac{5}{4}(0.720) = 0.90$$
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4Interpret the scale level (item mean = 3.63): 1.00–2.40 = Low | 2.41–3.40 = Moderate | 3.41–5.00 = High → Job satisfaction is HIGH.
18.13
Total mean (/25)
3.63
Item mean (/5)
0.90
Cronbach α
HIGH
Satisfaction level
APA-7
A 5-item job satisfaction scale (Likert, 1–5) demonstrated excellent internal consistency, α = .90. Participants reported high job satisfaction (M = 3.63, SD = 0.87, out of 5), with item means ranging from 3.50 to 3.88.