A study based on a sample of size 25reported

What is 45/1000 reduced?

Anna35 [415]

9/200. divide 45 and 1000 both by 5 to reduce it.

8 0

3 years ago

Roger and Mary have an online account for buying movies. Roger's account balance is -$25

padilas [110]

Answer:

Rodgers simply because owing 25 dollars is more than 10

5 0

3 years ago

Missing value 0.003 x what = 3

Tomtit [17]

2x=3
so substitute y=mx+c
rise/run

3 0

3 years ago

Principal, $2000; Annual interest rate, 5.3%; time, 2 years<br> After 2 years?

dangina [55]

Answer:

212 and I= 2000+212

=2212

Step-by-step explanation:

3 0

3 years ago

As part of the Pew Internet and American Life Project, researchers conducted two surveys in late 2009. The first survey asked a

REY [17]

Answer:

The 95% confidence interval for the difference between the proportion of all U.S. teens and adults who use social networking sites is (0.223, 0.297). This means that we are 95% sure that the true difference of the proportion is in this interval, between 0.223 and 0.297.

Step-by-step explanation:

Before building the confidence interval we need to understand the central limit theorem and the subtraction of normal variables.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

Subtraction between normal variables:

When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.

Sample of 800 teens. 73% said that they use social networking sites.

This means that:

$p_T = 0.73, s_T = \sqrt{\frac{0.73*0.27}{800}} = 0.0157$

Sample of 2253 adults. 47% said that they use social networking sites.

This means that:

$p_A = 0.47,s_A = \sqrt{\frac{0.47*0.53}{2253}} = 0.0105$

Distribution of the difference:

$p = p_T - p_A = 0.73 - 0.47 = 0.26$

$s = \sqrt{s_T^2+s_A^2} = \sqrt{0.0157^2+0.0105^2} = 0.019$

Confidence interval:

Is given by:

$p \pm zs$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

Lower bound:

$p - 1.96s = 0.26 - 1.96*0.019 = 0.223$

Upper bound:

$p + 1.96s = 0.26 + 1.96*0.019 = 0.297$

The 95% confidence interval for the difference between the proportion of all U.S. teens and adults who use social networking sites is (0.223, 0.297). This means that we are 95% sure that the true difference of the proportion is in this interval, between 0.223 and 0.297.

3 0

3 years ago