Verify the identity.<br> tan^5x = tan^3xsec^2x -tan^3x

A magazine reports that women trust recommendations from a particular social networking site more than recommendations from any

Misha Larkins [42]

Answer:

a) $p_A$ represent the real population proportion for women

$\hat p_A =\frac{123}{150}=0.82$ represent the estimated proportion for women

$n_A=150$ is the sample size selected for women

b) $p_B$ represent the real population proportion for male

$\hat p_B =\frac{102}{170}=0.6$ represent the estimated proportion for male

$n_B=170$ is the sample size selected for male

c) $(0.82-0.6) - 1.96 \sqrt{\frac{0.82(1-0.82)}{150} +\frac{0.6(1-0.6)}{170}}=0.124$

$(0.82-0.6) + 1.96 \sqrt{\frac{0.82(1-0.82)}{150} +\frac{0.6(1-0.6)}{170}}=0.316$

And the 95% confidence interval would be given (0.124;0.316).

We are confident at 95% that the difference between the two proportions is between $0.124 \leq p_A -p_B \leq 0.316$

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".

Part a

$p_A$ represent the real population proportion for women

$\hat p_A =\frac{123}{150}=0.82$ represent the estimated proportion for women

$n_A=150$ is the sample size selected for women

Part b

$p_B$ represent the real population proportion for male

$\hat p_B =\frac{102}{170}=0.6$ represent the estimated proportion for male

$n_B=170$ is the sample size elected for male

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

Part c

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 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.