Mathematics, 19.12.2019 20:31 bks53

In march 2015, the public policy institute of california (ppic) surveyed 7525 likely voters living in california. this is the 148th ppic research poll, and is part of a survey series that started in 1998. ppic researchers find that 3266 survey participants are registered democrats and 2137 survey participants are registered republicans. ppic is interested in the difference between the proportion of registered democrats and the proportion of registered republicans in california. ppic researchers calculate that the standard error for the proportion of registered democrats minus registered republicans is about 0.008. of the 3266 registered democrats, 1894 approve of the way the california legislature is handling its job. of the 2137 registered republicans, 385 approve of the way the california legislature is handling its job. what is the 90% confidence interval to estimate the difference in approval for the california legislature based on political party affiliation?

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Answer from: jwbri

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

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.

$z_{\alpha/2}=1.64$

And replacing into the confidence interval formula we got:

$(0.580-0.180) - 1.64 \sqrt{\frac{0.580(1-0.580)}{3266} +\frac{0.180(1-0.180)}{2137}}=0.380$

$(0.580-0.180) + 1.64 \sqrt{\frac{0.580(1-0.580)}{3266} +\frac{0.180(1-0.180)}{2137}}=0.420$

And the 99% confidence interval would be given (0.380;0.420).

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

Answer from: Quest

answer: the correct one is a

step-by-step explanation:

Answer from: Quest

answer: procent 54 wynosi 33,33%

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