HALP me please, mathematics
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Answer:
The 90% confidence interval for the difference between proportions is (-0.260, -0.165).
Step-by-step explanation:
<em>The question is incomplete. The complete question is:</em>
<em>"A survey reported in Time magazine included the question ‘‘Do you favor a federal law requiring a 15 day waiting period to purchase a gun?” Results from a random sample of US citizens showed that </em><em>318 of the 520 men </em><em>who were surveyed supported this proposed law while </em><em>379 of the 460 women </em><em>sampled said ‘‘yes”. Use this information to find a </em><em>90% confidence interval</em><em> for the difference in the two proportions, </em><em>pm - pw.</em><em> </em><em>Subscript pm</em><em> is the proportion of men who support the proposed law and </em><em>pw</em><em> is the proportion of women who support the proposed law. (Round answers to 3 decimal places.)"</em>
We want to calculate the bounds of a 90% confidence interval.
For a 90% CI, the critical value for z is z=1.645.
The sample of men, of size nm=-0.26 has a proportion of pm=0.612.
The sample 2, of size nw= has a proportion of pw=0.824.
The difference between proportions is (pm-pw)=-0.212.
The pooled proportion, needed to calculate the standard error, is:
The estimated standard error of the difference between means is computed using the formula:
Then, the margin of error is:
Then, the lower and upper bounds of the confidence interval are:
The 90% confidence interval for the difference between proportions is (-0.260, -0.165).