document.write( "Question 1171066: The performance of 50 Psychology students showed a mean of 98 with a standard deviation of 9.2. A sample of 40 students showed an average performance of 95 with a standard deviation of 7.3. Is there a difference in performance between the two samples using 0.05 level, two-tailed? \n" ); document.write( "
Algebra.Com's Answer #850994 by CPhill(1987) You can put this solution on YOUR website! Here's how to conduct the hypothesis test to determine if there's a difference in performance between the two samples.\r \n" ); document.write( "\n" ); document.write( "**1. State the Hypotheses**\r \n" ); document.write( "\n" ); document.write( "* **Null Hypothesis (H₀):** There is no difference in the mean performance between the two samples. \n" ); document.write( " * H₀: μ₁ = μ₂ \n" ); document.write( "* **Alternative Hypothesis (H₁):** There is a difference in the mean performance between the two samples. \n" ); document.write( " * H₁: μ₁ ≠ μ₂ (two-tailed test)\r \n" ); document.write( "\n" ); document.write( "**2. Determine the Test Statistic**\r \n" ); document.write( "\n" ); document.write( "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.)\r \n" ); document.write( "\n" ); document.write( "The formula for the z-statistic is:\r \n" ); document.write( "\n" ); document.write( "z = (x̄₁ - x̄₂) / √( (σ₁²/n₁) + (σ₂²/n₂) )\r \n" ); document.write( "\n" ); document.write( "Where:\r \n" ); document.write( "\n" ); document.write( "* x̄₁ = mean of sample 1 \n" ); document.write( "* x̄₂ = mean of sample 2 \n" ); document.write( "* σ₁ = standard deviation of sample 1 \n" ); document.write( "* σ₂ = standard deviation of sample 2 \n" ); document.write( "* n₁ = sample size of sample 1 \n" ); document.write( "* n₂ = sample size of sample 2\r \n" ); document.write( "\n" ); document.write( "**3. Calculate the Test Statistic**\r \n" ); document.write( "\n" ); document.write( "* x̄₁ = 98 \n" ); document.write( "* σ₁ = 9.2 \n" ); document.write( "* n₁ = 50 \n" ); document.write( "* x̄₂ = 95 \n" ); document.write( "* σ₂ = 7.3 \n" ); document.write( "* n₂ = 40\r \n" ); document.write( "\n" ); document.write( "z = (98 - 95) / √((9.2²/50) + (7.3²/40)) \n" ); document.write( "z = 3 / √((84.64/50) + (53.29/40)) \n" ); document.write( "z = 3 / √(1.6928 + 1.33225) \n" ); document.write( "z = 3 / √3.02505 \n" ); document.write( "z = 3 / 1.739 \n" ); document.write( "z ≈ 1.725\r \n" ); document.write( "\n" ); document.write( "**4. Determine the Critical Value or P-value**\r \n" ); document.write( "\n" ); document.write( "* Significance level (α) = 0.05 \n" ); document.write( "* Type of test: Two-tailed\r \n" ); document.write( "\n" ); document.write( "* **Critical Value Approach:** \n" ); document.write( " * For a two-tailed test with α = 0.05, the critical z-values are ±zα/2 = ±z0.025. \n" ); document.write( " * Using a standard normal distribution table or a calculator, we find that z0.025 ≈ 1.96.\r \n" ); document.write( "\n" ); document.write( "* **P-value Approach:** \n" ); document.write( " * Using a z-table or calculator, we find the p-value associated with z = 1.725 for a two-tailed test. \n" ); document.write( " * P(Z > 1.725) ≈ 0.0423 \n" ); document.write( " * P(Z < -1.725) ≈ 0.0423 \n" ); document.write( " * p-value = 2 \* 0.0423 ≈ 0.0846\r \n" ); document.write( "\n" ); document.write( "**5. Make a Decision**\r \n" ); document.write( "\n" ); document.write( "* **Critical Value Approach:** \n" ); document.write( " * The calculated z-statistic is 1.725. \n" ); document.write( " * The critical z-values are ±1.96. \n" ); document.write( " * Since |1.725| < 1.96, we fail to reject the null hypothesis.\r \n" ); document.write( "\n" ); document.write( "* **P-value Approach:** \n" ); document.write( " * The p-value is approximately 0.0846. \n" ); document.write( " * The significance level is 0.05. \n" ); document.write( " * Since the p-value (0.0846) is greater than the significance level (0.05), we fail to reject the null hypothesis.\r \n" ); document.write( "\n" ); document.write( "**6. Conclusion**\r \n" ); document.write( "\n" ); document.write( "There is not sufficient evidence at the 0.05 significance level to conclude that there is a difference in performance between the two samples. \n" ); document.write( " |