Question 1171066
Here's how to conduct the hypothesis test to determine if there's a difference in performance between the two samples.

**1. State the Hypotheses**

* **Null Hypothesis (H₀):** There is no difference in the mean performance between the two samples.
    * H₀: μ₁ = μ₂
* **Alternative Hypothesis (H₁):** There is a difference in the mean performance between the two samples.
    * H₁: μ₁ ≠ μ₂ (two-tailed test)

**2. Determine the Test Statistic**

Since we are comparing the means of two independent samples and we have sample standard deviations, we will use a two-sample z-test. (We can use a z-test because the sample sizes are reasonably large.)

The formula for the z-statistic is:

z = (x̄₁ - x̄₂) / √( (σ₁²/n₁) + (σ₂²/n₂) )

Where:

* x̄₁ = mean of sample 1
* x̄₂ = mean of sample 2
* σ₁ = standard deviation of sample 1
* σ₂ = standard deviation of sample 2
* n₁ = sample size of sample 1
* n₂ = sample size of sample 2

**3. Calculate the Test Statistic**

* x̄₁ = 98
* σ₁ = 9.2
* n₁ = 50
* x̄₂ = 95
* σ₂ = 7.3
* n₂ = 40

z = (98 - 95) / √((9.2²/50) + (7.3²/40))
z = 3 / √((84.64/50) + (53.29/40))
z = 3 / √(1.6928 + 1.33225)
z = 3 / √3.02505
z = 3 / 1.739
z ≈ 1.725

**4. Determine the Critical Value or P-value**

* Significance level (α) = 0.05
* Type of test: Two-tailed

* **Critical Value Approach:**
    * For a two-tailed test with α = 0.05, the critical z-values are ±z<sub>α/2</sub> = ±z<sub>0.025</sub>.
    * Using a standard normal distribution table or a calculator, we find that z<sub>0.025</sub> ≈ 1.96.

* **P-value Approach:**
    * Using a z-table or calculator, we find the p-value associated with z = 1.725 for a two-tailed test.
    * P(Z > 1.725) ≈ 0.0423
    * P(Z < -1.725) ≈ 0.0423
    * p-value = 2 \* 0.0423 ≈ 0.0846

**5. Make a Decision**

* **Critical Value Approach:**
    * The calculated z-statistic is 1.725.
    * The critical z-values are ±1.96.
    * Since |1.725| < 1.96, we fail to reject the null hypothesis.

* **P-value Approach:**
    * The p-value is approximately 0.0846.
    * The significance level is 0.05.
    * Since the p-value (0.0846) is greater than the significance level (0.05), we fail to reject the null hypothesis.

**6. Conclusion**

There is not sufficient evidence at the 0.05 significance level to conclude that there is a difference in performance between the two samples.