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:

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

Then, the margin of error is:

$MOE=z \cdot s_{p1-p2}=1.96\cdot 0.022=0.0438$

Then, the lower and upper bounds of the confidence interval are:

$LL=(p_1-p_2)-z\cdot s_{p1-p2} = -0.058-0.0438=-0.102\\\\UL=(p_1-p_2)+z\cdot s_{p1-p2}= -0.058+0.0438=-0.014$

The 95% confidence interval for the difference between proportions is (-0.102, -0.014).

7 0

3 years ago

Reduce this algebraic fraction. y^3 -3y^2+y-3 / y^2-9

blondinia [14]

Answer:

y³ - 3y² + 1/y+3

Step-by-step explanation:

y² - 9 = (y - 3) ( y + 3)

y-3/(y-3)(y+3) = 1/y+3

6 0

3 years ago

Read 2 more answers

{(-1, 2), (0, 2), (5, 2)} is a function.<br> O True<br> O False

jarptica [38.1K]

True. When the points are plotted on a graph and connected, they pass the vertical line test. No x value is the same.

6 0

3 years ago

Read 2 more answers

Suppose p" must approximate p with relative error at most 10-3 . Find the largest interval in which p* must lie for each value o

goblinko [34]

Answer:

$[p-|p|*10^{-3} \, , \, p+|p|* 10^-3]$

Step-by-step explanation

The relative error is the absolute error divided by the absolute value of p. for an approximation p*, the relative error is

r = |p*-p|/|p|

we want r to be at most 10⁻³, thus

|p*-p|/|p| ≤ 10⁻³

|p*-p| ≤ |p|* 10⁻³

therefore, p*-p should lie in the interval [ - |p| * 10⁻³ , |p| * 10⁻³ ], and as a consecuence, p* should be in the interval [p - |p| * 10⁻³ , p + |p| * 10⁻³ ]

8 0

3 years ago