document.write( "Question 1171313: Inference (Two Populations), Chi-Squared Tests
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document.write( "1. A study comparing children’s reading age (in months) was developed using identical twin
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document.write( "toddlers. One set of 6 twins played for 2 hours each day with educational toys
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document.write( "(experimental group), the corresponding set of 6 twins played for 2 hours each day with
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document.write( "non-educational toys (control group). The mean difference in reading age between the
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document.write( "experimental group and control group of 6 sets of twins was -2.33 months and the
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document.write( "standard deviation of the sample difference was 2.16 months. Set up a hypothesis test to
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document.write( "determine if there is a difference between the 2 groups and use the appropriate sample
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document.write( "test statistic to determine if the difference in reading age is significant at the 5% level \n" );
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Algebra.Com's Answer #850969 by CPhill(1959)![]() ![]() You can put this solution on YOUR website! Absolutely! Let's break down this hypothesis test step-by-step.\r \n" ); document.write( "\n" ); document.write( "**1. Set up the Hypotheses:**\r \n" ); document.write( "\n" ); document.write( "* **Null Hypothesis (H₀):** There is no difference in the mean reading age between the experimental group (educational toys) and the control group (non-educational toys). In other words, the mean difference is zero. \n" ); document.write( " * H₀: μd = 0 (where μd is the mean difference) \n" ); document.write( "* **Alternative Hypothesis (H₁):** There is a difference in the mean reading age between the two groups. \n" ); document.write( " * H₁: μd ≠ 0 (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 related groups (identical twins) and we have a small sample size (n = 6), we will use a paired t-test.\r \n" ); document.write( "\n" ); document.write( "The formula for the t-statistic is:\r \n" ); document.write( "\n" ); document.write( "t = (mean difference) / (standard deviation of the difference / √sample size)\r \n" ); document.write( "\n" ); document.write( "t = μd / (sd / √n)\r \n" ); document.write( "\n" ); document.write( "Where:\r \n" ); document.write( "\n" ); document.write( "* μd = -2.33 months (mean difference) \n" ); document.write( "* sd = 2.16 months (standard deviation of the difference) \n" ); document.write( "* n = 6 (number of twin pairs)\r \n" ); document.write( "\n" ); document.write( "**3. Calculate the Test Statistic:**\r \n" ); document.write( "\n" ); document.write( "t = -2.33 / (2.16 / √6) \n" ); document.write( "t = -2.33 / (2.16 / 2.449) \n" ); document.write( "t = -2.33 / 0.882 \n" ); document.write( "t ≈ -2.642\r \n" ); document.write( "\n" ); document.write( "**4. Determine the Degrees of Freedom:**\r \n" ); document.write( "\n" ); document.write( "Degrees of freedom (df) = n - 1 \n" ); document.write( "df = 6 - 1 = 5\r \n" ); document.write( "\n" ); document.write( "**5. Determine the Critical Value or P-value:**\r \n" ); document.write( "\n" ); document.write( "* **P-value Approach:** \n" ); document.write( " * Using a t-distribution table or a calculator, we find the p-value associated with a t-statistic of -2.642 and 5 degrees of freedom. \n" ); document.write( " * Since this is a two-tailed test, we look for the probability of observing a t-statistic as extreme as -2.642 or 2.642. \n" ); document.write( " * The p-value is approximately 0.0459.\r \n" ); document.write( "\n" ); document.write( "* **Critical Value Approach:** \n" ); document.write( " * For a two-tailed test with α = 0.05 and df = 5, the critical t-values are ±2.571.\r \n" ); document.write( "\n" ); document.write( "**6. Make a Decision:**\r \n" ); document.write( "\n" ); document.write( "* **P-value Approach:** \n" ); document.write( " * Since the p-value (0.0459) is less than the significance level (0.05), we reject the null hypothesis.\r \n" ); document.write( "\n" ); document.write( "* **Critical Value Approach:** \n" ); document.write( " * Since the calculated t-statistic (-2.642) is less than the critical value (-2.571), we reject the null hypothesis.\r \n" ); document.write( "\n" ); document.write( "**7. Conclusion:**\r \n" ); document.write( "\n" ); document.write( "There is sufficient evidence to conclude that there is a statistically significant difference in the mean reading age between children who play with educational toys and those who play with non-educational toys at the 5% significance level. \n" ); document.write( " \n" ); document.write( " |