**Question 3**On June 28, 2012, the U.S. Supreme Court upheld the much debated 2010 healthcare law, declaring it constitutional. A Gallup poll released the day after this decision indicates that 46% of 1,012 Americans agree with this decision. At a 95% confidence level, this sample has a 3% margin of error.(a) Based on this information, determine if the following statements are true or false, and explain your reasoning.1. We are 95% confident that between 43% and 49% of Americans in this sample support the decision of the U.S. Supreme Court on the 2010 healthcare law.2. (43%, 49%) is the 95% confidence interval for \( p \), the proportion of Americans support the decision.3. We are 95% confident that between 43% and 49% of Americans support the decision of the U.S. Supreme Court on the 2010 healthcare law.4. If we considered many random samples of 1,012 Americans, and we calculated the sample proportions of those who support the decision of the U.S. Supreme Court, 95% of those sample proportions will be between 43% and 49%.5. The margin of error at a 90% confidence level would be higher than 3%.

Given Information: A Gallup poll released the day after this decision indicates that 46% of 1012 Americans agree with this decision. Confidence level is 95% Margin of error is 3% (a) (1) We are 95% confident that between 43% and 49% of Americans in this sample support the decision of the U.S supreme court on the 2010 healthcare law. A confidence interval indicates a range of values in which the population parameter is expected to lie. This statement is false, because the confidence interval is used to estimate the true population parameter and not the sample. (2) Sample proportion is p^=0.46 The 95% confidence interval for population proportion is given by the formula: C.I=p^±zα/2p^1-p^n where, zα/2 is the z-critical value at 1-0.952=0.025 significance level. Using the standard normal table, look for 0.025 in the entries of the table, the z-value corresponding to it is -1.96 Thus, the z-critical value is 1.96 (ignore minus sign). Substitute the values in the formula: C.I=p^±zα/2p^1-p^n=0.46±1.960.461-0.461012=0.46±1.96*0.01566698904=0.4292927015, 0.4907072985≈0.43, 0.49≈43%, 49% The 95% confidence interval for the true proportion p is (0.43, 0.49) Hence, the statement is true. (3) We are 95% confident that between 43% and 49% of Americans supported the decision of the U.S supreme court on the 2010 healthcare law. This statement is true because this is the interpretation of the confidence interval. (4) The given statement is false, because a 95% confidence interval for population proportion means that the confidence interval will contain the population proportion 95% of the time but might not be the same range (43%, 49%). (5) The margin of error at a 90% confidence level would be higher than 3%. The margin of error for 95% confidence level is: E=zα/2p^1-p^n=1.960.461-0.461012=1.96*0.01566698904=0.03070729852≈0.03≈3% The margin of error for 90% confidence level is: E=zα/2p^1-p^n=z0.10/2p^1-p^n=z0.05p^1-p^n=1.6450.461-0.461012 [Use Excel function "=NORM.S.INV(0.05)"]=1.645*0.01566698904=0.02577219697≈0.0257≈2.57% Since the margin of error for 90% confidence level (2.57%) is less than the margin of error for 95% confidence level (3%), the given statement is false.