Question

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 p̂'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)	p̂ = (x/n)
Texas	689	1079	0.63855
New York	455	784	0.58036

Standard Error (SE) = 0.02285

The Test: Test the claim that the proportion of gun murders was significantly greater in Texas than New York in 2011. Use a 0.01 significance level.

(a) Letting p̂₁ be the proportion of gun murders in Texas and p̂₂ be the proportion from New York, calculate the test statistic using software or the formulaz =

(p̂1 − p̂2) − δp

SE

where δ_p 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?

reject H₀

fail to reject H₀    

(d) Choose the appropriate concluding statement.

The data supports the claim that the proportion of gun murders was significantly greater in Texas than New York.

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.

Answer 1