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Gun Murders - Texas vs New York - Significance Test In 2011, New York had much stricter gun laws than Texas. For that year, the proportion of gun murders in Texas was greater than in New York. Here we test whether or not the proportion was significantly greater in Texas. The table below gives relevant information. Here, the p's are population proportions but you should treat them as sample proportions. The standard error (SE) is given to save calculation time if you are not using software. Data Summary number of total number Proportion State gun murders (x) of murders (n) p = (x) Texas 689 1089 0.63269 New York 445 774 0.57494 Standard Error (SE) = 0.02294 The Test: Test the claim that the proportion of gun murders was significantly greater in Texas than New York in 2011. Use a 0.05 significance level. (a) Letting p1 be the proportion of gun murders in Texas and p2 be the proportion from New York, calculate the test statistic using software or the formula - (P1 - P2) - 5p SE where op is the hypothesized difference in proportions from the null hypothesis and the standard error (SE) is given with the data. Round your answer to 2 decimal places. z = To account for hand calculations -vs- software, your answer must be within 0.01 of the true answer. (b) Use software or the z-table to get the P-value of the test statistic. Round to 4 decimal places. P-value = (c) What is the conclusion regarding the null hypothesis? O reject Ho O fail to reject Ho 'd) Choose the appropriate concluding statement. O The data supports the claim that the proportion of gun murders was significantly greater in Texas than New York. O While the proportion of gun murders in Texas was greater than New York, the difference was not great enough to be considered significant. We have proven that the stricter gun laws in New York actually decreased the proportion of gun murders below the rate in Texas. We have proven there was no difference in the proportion of gun murders between Texas and New York