document.write( "Question 1193999: A lab technician is tested for her consistency by taking multiple measurements of cholesterol levels from the same blood sample. The target accuracy is a standard deviation less than 1.2. If 20 measurements are taken and their standard deviation is 1.1, is there enough evidence at a=.05 that she is meeting the target accuracy?\r
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document.write( "The correct Null and Alternative Hypothesis statement would be?\r
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document.write( "The correct critical value would be?\r
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document.write( "The correct test value would be?\r
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document.write( "The decision and summary statement would be?
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Algebra.Com's Answer #848535 by parmen(42) ![]() You can put this solution on YOUR website! **1. Set up Hypotheses**\r \n" ); document.write( "\n" ); document.write( "* **Null Hypothesis (H₀):** σ² ≥ 1.2² (The population variance of the technician's measurements is greater than or equal to 1.44) \n" ); document.write( "* **Alternative Hypothesis (H₁):** σ² < 1.2² (The population variance of the technician's measurements is less than 1.44)\r \n" ); document.write( "\n" ); document.write( "**2. Determine the Test Statistic**\r \n" ); document.write( "\n" ); document.write( "* We will use the Chi-Square test statistic for testing the population variance. \n" ); document.write( "* **Test Statistic:** χ² = (n - 1) * s² / σ₀² \n" ); document.write( " * where: \n" ); document.write( " * n = sample size (20 measurements) \n" ); document.write( " * s² = sample variance (1.1² = 1.21) \n" ); document.write( " * σ₀² = hypothesized population variance (1.2² = 1.44)\r \n" ); document.write( "\n" ); document.write( "* **Calculate the Test Statistic:** \n" ); document.write( " * χ² = (20 - 1) * 1.21 / 1.44 = 19 * 1.21 / 1.44 = 15.96\r \n" ); document.write( "\n" ); document.write( "**3. Determine the Critical Value**\r \n" ); document.write( "\n" ); document.write( "* **Degrees of Freedom (df):** df = n - 1 = 20 - 1 = 19 \n" ); document.write( "* **Significance Level (α):** α = 0.05 \n" ); document.write( "* **Critical Value:** \n" ); document.write( " * Since this is a left-tailed test, we find the critical value from the Chi-Square distribution table with 19 degrees of freedom and an area of 0.05 in the left tail. \n" ); document.write( " * Using a Chi-Square table or statistical software, the critical value for χ²(0.05, 19) is approximately 8.907.\r \n" ); document.write( "\n" ); document.write( "**4. Decision**\r \n" ); document.write( "\n" ); document.write( "* **Compare Test Statistic to Critical Value:** \n" ); document.write( " * Our calculated test statistic (15.96) is greater than the critical value (8.907).\r \n" ); document.write( "\n" ); document.write( "* **Decision:** \n" ); document.write( " * Since the test statistic does not fall in the rejection region (less than the critical value), we **fail to reject the null hypothesis**.\r \n" ); document.write( "\n" ); document.write( "**5. Summary Statement**\r \n" ); document.write( "\n" ); document.write( "* There is **not enough evidence** at the 0.05 significance level to conclude that the lab technician is meeting the target accuracy of a standard deviation less than 1.2. \r \n" ); document.write( "\n" ); document.write( "**Key Points:**\r \n" ); document.write( "\n" ); document.write( "* We use the Chi-Square test for testing hypotheses about population variance. \n" ); document.write( "* The critical value is determined based on the significance level and degrees of freedom. \n" ); document.write( "* We compare the test statistic to the critical value to make a decision about the null hypothesis. \n" ); document.write( " \n" ); document.write( " |