63x^3y+105x^2y divided by -7xy help, please

Mathematics

1 answer:

STALIN [3.7K]3 years ago

7 0

Hope this helps, now you know the answer and how to do it. HAVE A BLESSED AND WONDERFUL DAY! As well as a great Valentines Day! :-)

- Cutiepatutie ☺❀❤

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A College Alcohol Study has interviewed random samples of students at four-year colleges. In the most recent study, 494 of 1000

Vinil7 [7]

Answer:

The 95% confidence interval for the difference between the proportion of women who drink alcohol and the proportion of men who drink alcohol is (-0.102, -0.014) or (-10.2%, -1.4%).

Step-by-step explanation:

We want to calculate the bounds of a 95% confidence interval of the difference between proportions.

For a 95% CI, the critical value for z is z=1.96.

The sample 1 (women), of size n1=1000 has a proportion of p1=0.494.

$p_1=X_1/n_1=494/1000=0.494$

The sample 2 (men), of size n2=1000 has a proportion of p2=0.552.

$p_2=X_2/n_2=552/1000=0.552$

The difference between proportions is (p1-p2)=-0.058.

$p_d=p_1-p_2=0.494-0.552=-0.058$

The pooled proportion, needed to calculate the standard error, is:

$p=\dfrac{X_1+X_2}{n_1+n_2}=\dfrac{494+552}{1000+1000}=\dfrac{1046}{2000}=0.523$

The estimated standard error of the difference between means is computed using the formula:

$s_{p1-p2}=\sqrt{\dfrac{p(1-p)}{n_1}+\dfrac{p(1-p)}{n_2}}=\sqrt{\dfrac{0.523*0.477}{1000}+\dfrac{0.523*0.477}{1000}}\\\\\\s_{p1-p2}=\sqrt{0.000249+0.000249}=\sqrt{0.000499}=0.022$