Pls help :) would mean a lot

nlexa [21]

X hope that I helped you

7 0

3 years ago

An investor has an account with stock from two different companies. Last year, her stock in Company A was worth $5750 and her st

V125BC [204]

Answer:13.6%

Step-by-step explanation: I got the same problem and pressed give up.

8 0

3 years ago

A package of pencils cost $6.25 per set at an office supply store. Which equation can be used to determine the price, P, of n se

ladessa [460]

To solve, it would be 6.25n. That eliminates both 6n25 over P, P equals 6.25 over n, and n=6.25P, leaving P=6.25n as the answer.

5 0

3 years ago

Read 2 more answers

In March 2015, the Public Policy Institute of California (PPIC) surveyed 7525 likely voters living in California. This is the 14

lbvjy [14]

Answer:

We are confident at 99% that the difference between the two proportions is between $0.380 \leq p_{Republicans} -p_{Democrats} \leq 0.420$

Step-by-step explanation:

Part a

Data given and notation

$X_{D}=3266$ represent the number people registered as Democrats

$X_{R}=2137$ represent the number of people registered as Republicans

$n=7525$ sampleselcted

$\hat p_{D}=\frac{3266}{7525}=0.434$ represent the proportion of people registered as Democrats

$\hat p_{R}=\frac{2137}{7525}=0.284$ represent the proportion of people registered as Republicans

The standard error is given by this formula:

$SE=\sqrt{\frac{\hat p_D (1-\hat p_D)}{n_{D}}+\frac{\hat p_R (1-\hat p_R)}{n_{R}}}$

And the standard error estimated given by the problem is 0.008

Part b

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$p_A$ represent the real population proportion of Democrats that approve of the way the California Legislature is handling its job

$\hat p_A =\frac{1894}{3266}=0.580$ represent the estimated proportion of Democrats that approve of the way the California Legislature is handling its job

$n_A=3266$ is the sample size for Democrats

$p_B$ represent the real population proportion of Republicans that approve of the way the California Legislature is handling its job

$\hat p_B =\frac{385}{2137}=0.180$ represent the estimated proportion of Republicans that approve of the way the California Legislature is handling its job

$n_B=2137$ is the sample for Republicans

$z$ represent the critical value for the margin of error

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

The confidence interval for the difference of two proportions would be given by this formula

$(\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}$

For the 90% confidence interval the value of $\alpha=1-0.90=0.1$ and $\alpha/2=0.05$ , with that value we can find the quantile required for the interval in the normal standard distribution.