PLEASE HELPP

1 answer:

3 0

$-6\cdot q=24p$
$q=-4$

$24\cdot (-4) =?$
$-96=?$

Let's check it:
$-96\cdot (-4)=384$
$384=384$
$L=R$

The answer: C

<em>:)</em>

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a 2015 Gallup poll of 1,627 adults found that only 22% felt fully engaged with their mortgage provider ?

Gala2k [10]

Answer:

1. The population of interest on this case is people who felt fully engaged with their mortgage provider

2. On this case the sample size is 1627 the number of people sample selected.

3. $ME=1.96\sqrt{\frac{0.22 (1-0.22)}{1627}}=0.0201$

4. The 95% confidence interval would be given (0.1999;0.2401).

5. We are confident 95% that the true proportion of proportion is between 0.1999 and 0.2401.

Step-by-step explanation:

Previous concepts

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".

The population proportion have the following distribution

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

1. What is the population?

The population of interest on this case is people who felt fully engaged with their mortgage provider

2. What is the sample size?

On this case the sample size is 1627 the number of people sample selected.

3. What is the 95% margin of error?

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by:

$z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96$

The margin of error for the proportion interval is given by this formula:

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)

If we replace we got:

$ME=1.96\sqrt{\frac{0.22 (1-0.22)}{1627}}=0.0201$

4. What is the 95% confidence interval?

The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.