SOLUTION: Data is collected from 30 male and 20 female employees of SMIU. 50% of the female scored 4 and remaining 2; 30% of the males scored 3, 30% scored 4 and remaining scored 5. Sd of fe

Question 1174091: Data is collected from 30 male and 20 female employees of SMIU. 50% of the female scored 4 and remaining 2; 30% of the males scored 3, 30% scored 4 and remaining scored 5. Sd of females is 0.8 and that of males is .8. Design hypothesis and test it?
Answer by CPhill(1987) (Show Source): You can put this solution on YOUR website!
Let's break down this problem and design/test a hypothesis.
**1. Data Organization**
* **Females:**
* Total: 20
* 50% scored 4: 20 * 0.50 = 10 females scored 4.
* 50% scored 2: 20 * 0.50 = 10 females scored 2.
* Female scores: 10 with 4, 10 with 2.
* Sd of females = 0.8
* **Males:**
* Total: 30
* 30% scored 3: 30 * 0.30 = 9 males scored 3.
* 30% scored 4: 30 * 0.30 = 9 males scored 4.
* 40% scored 5: 30 * 0.40 = 12 males scored 5.
* Male scores: 9 with 3, 9 with 4, 12 with 5.
* Sd of males = 0.8
**2. Calculate Means**
* **Female Mean (μf):**
* (10 * 4 + 10 * 2) / 20 = (40 + 20) / 20 = 60 / 20 = 3
* **Male Mean (μm):**
* (9 * 3 + 9 * 4 + 12 * 5) / 30 = (27 + 36 + 60) / 30 = 123 / 30 = 4.1
**3. Hypothesis Design**
* We want to test if there's a significant difference between the mean scores of male and female employees.
* **Null Hypothesis (H0):** There is no significant difference between the mean scores of male and female employees. (μm = μf)
* **Alternative Hypothesis (H1):** There is a significant difference between the mean scores of male and female employees. (μm ≠ μf)
* This is a two-tailed test.
**4. Test Statistic**
* We'll use a two-sample t-test because we're comparing two means. Since we have the standard deviation of the populations we can use a z test.
* We will use the following formula.
* z = (μm - μf) / sqrt((σm^2/nm) + (σf^2/nf))
* Where:
* μm = 4.1 (male mean)
* μf = 3 (female mean)
* σm = 0.8 (male standard deviation)
* σf = 0.8 (female standard deviation)
* nm = 30 (number of males)
* nf = 20 (number of females)
**5. Calculation**
* z = (4.1 - 3) / sqrt((0.8^2/30) + (0.8^2/20))
* z = 1.1 / sqrt((0.64/30) + (0.64/20))
* z = 1.1 / sqrt(0.02133 + 0.032)
* z = 1.1 / sqrt(0.05333)
* z = 1.1 / 0.2309
* z ≈ 4.76
**6. Determine the Critical Value and Make a Decision**
* Let's use a significance level (alpha) of 0.05.
* For a two-tailed test at alpha = 0.05, the critical z-values are ±1.96.
* Our calculated z-value (4.76) is far outside this range.
* Therefore, we reject the null hypothesis.
**7. Conclusion**
* There is a statistically significant difference between the mean scores of male and female employees at SMIU.
* The male employees, on average, scored higher than the female employees.

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