document.write( "Question 1186605: 6. A school administrator claims that less than 50% of the students of the school are dissatisfied by the community cafeteria service. Test this claim by using sample data obtained from a survey of 500 students of the school where 54% indicated their dissatisfaction of the community cafeteria service. Use alpha = 0.05\r
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document.write( "answer the problem using the steps in hypothesis testing. \n" );
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Algebra.Com's Answer #849423 by CPhill(1987) You can put this solution on YOUR website! Here's how to conduct a hypothesis test for the school administrator's claim:\r \n" ); document.write( "\n" ); document.write( "**1. State the Hypotheses:**\r \n" ); document.write( "\n" ); document.write( "* **Null Hypothesis (H₀):** p ≥ 0.50 (The proportion of dissatisfied students is greater than or equal to 50%) \n" ); document.write( "* **Alternative Hypothesis (H₁ or Ha):** p < 0.50 (The proportion of dissatisfied students is less than 50%) This is a *left-tailed test*.\r \n" ); document.write( "\n" ); document.write( "**2. Determine the Level of Significance (alpha):**\r \n" ); document.write( "\n" ); document.write( "* α = 0.05 (This is given in the problem.)\r \n" ); document.write( "\n" ); document.write( "**3. Calculate the Test Statistic:**\r \n" ); document.write( "\n" ); document.write( "* **Sample Proportion (p̂):** p̂ = 0.54 (54% of the 500 students surveyed) \n" ); document.write( "* **Hypothesized Proportion (p₀):** p₀ = 0.50 (The proportion under the null hypothesis) \n" ); document.write( "* **Standard Error (SE):** SE = sqrt[(p₀ * (1 - p₀)) / n] = sqrt[(0.50 * (1 - 0.50)) / 500] = sqrt(0.0005) ≈ 0.0224 \n" ); document.write( "* **Z-score:** z = (p̂ - p₀) / SE = (0.54 - 0.50) / 0.0224 ≈ 1.79\r \n" ); document.write( "\n" ); document.write( "**4. Determine the Critical Value or the P-value:**\r \n" ); document.write( "\n" ); document.write( "* **Critical Value (Left-tailed test):** For α = 0.05, the critical z-value is -1.645. (You can find this using a z-table or a calculator.) \n" ); document.write( "* **P-value:** Since this is a left-tailed test, the p-value is the probability of observing a z-score as extreme as, or more extreme than, 1.79. Using a z-table or calculator, we find the area to the *left* of z = 1.79. Because this is the opposite of the tail we are interested in, we subtract this from 1. So p-value = 1 - 0.9633 = 0.0367.\r \n" ); document.write( "\n" ); document.write( "**5. Make a Decision:**\r \n" ); document.write( "\n" ); document.write( "* **Using Critical Values:** Our calculated z-score (1.79) is *greater* than the critical z-value (-1.645). Therefore, we fail to reject the null hypothesis. \n" ); document.write( "* **Using P-value:** The p-value (0.0367) is *less* than the level of significance (0.05). Because this is a left tailed test, we are interested in the values to the left of the critical value and therefore, reject the null hypothesis.\r \n" ); document.write( "\n" ); document.write( "**6. Conclusion:**\r \n" ); document.write( "\n" ); document.write( "There *is* sufficient evidence at the 0.05 level of significance to support the school administrator's claim that less than 50% of students are dissatisfied with the cafeteria service. \n" ); document.write( " \n" ); document.write( " |