document.write( "Question 1171061: A standardized test was administered to a thousand pupils with a mean average score of 85 and a standard deviation of 8. A random sample of 50 pupils was given the same test and showed an average score of 83.20. Is there evidence to show that this group has a lower performance than the ones in general at 0.05 level (two-tailed)? \n" ); document.write( "
Algebra.Com's Answer #850997 by CPhill(1959)\"\" \"About 
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Let's conduct the hypothesis test to see if the sample group has a lower performance than the general population.\r
\n" ); document.write( "\n" ); document.write( "**1. State the Hypotheses**\r
\n" ); document.write( "\n" ); document.write( "* **Null Hypothesis (H₀):** The sample group's mean score is equal to the population mean score.
\n" ); document.write( " * H₀: μ = 85
\n" ); document.write( "* **Alternative Hypothesis (H₁):** The sample group's mean score is different from the population mean score.
\n" ); document.write( " * H₁: μ ≠ 85 (two-tailed test)\r
\n" ); document.write( "\n" ); document.write( "**2. Determine the Test Statistic**\r
\n" ); document.write( "\n" ); document.write( "* We are given the population standard deviation, so we will use a z-test.
\n" ); document.write( "* The formula for the z-statistic is:\r
\n" ); document.write( "\n" ); document.write( " z = (x̄ - μ) / (σ / √n)\r
\n" ); document.write( "\n" ); document.write( " Where:
\n" ); document.write( " * x̄ = sample mean
\n" ); document.write( " * μ = population mean
\n" ); document.write( " * σ = population standard deviation
\n" ); document.write( " * n = sample size\r
\n" ); document.write( "\n" ); document.write( "**3. Calculate the Test Statistic**\r
\n" ); document.write( "\n" ); document.write( "* x̄ = 83.20
\n" ); document.write( "* μ = 85
\n" ); document.write( "* σ = 8
\n" ); document.write( "* n = 50\r
\n" ); document.write( "\n" ); document.write( " z = (83.20 - 85) / (8 / √50)
\n" ); document.write( " z = -1.8 / (8 / 7.071)
\n" ); document.write( " z = -1.8 / 1.131
\n" ); document.write( " z ≈ -1.591\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 = -1.591 for a two-tailed test.
\n" ); document.write( " * P(Z < -1.591) ≈ 0.0558
\n" ); document.write( " * P(Z > 1.591) ≈ 0.0558
\n" ); document.write( " * p-value = 2 \* 0.0558 ≈ 0.1116\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 -1.591.
\n" ); document.write( " * The critical z-values are ±1.96.
\n" ); document.write( " * Since |-1.591| < 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 0.1116.
\n" ); document.write( " * The significance level is 0.05.
\n" ); document.write( " * Since the p-value (0.1116) 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 the sample group has a lower performance than the general population.
\n" ); document.write( " \n" ); document.write( "
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