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2 years ago

Not sure how to solve this one

Mathematics

1 answer:

Alex73 [517]2 years ago

4 0

Answer:

$x\approx 63.4^0$

Step-by-step explanation:

In the given right triangle,

Measure of angle = x°

Opposite side of the angle = 12 units

Adjacent side of the angle = 6 units

By taking tangent ratio of the given angle,

tan(x°) = $\frac{\text{Opposite side}}{\text{Adjacent side}}$

= $\frac{12}{6}$

= 2

$x^0=\text{tan}^{-1}(2)}$

$x=63.43^0$

$x\approx 63.4^0$

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a) $\hat p=\frac{471}{1024}=0.460$

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$SE= \sqrt{\frac{\hat p(1-\hat p)}{n}}=\sqrt{\frac{0.460(1-0.460)}{1024}}=0.0156$

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b) The 99% confidence interval would be given by (0.429;0.491)

Step-by-step explanation:

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

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

Data given and notation

n=1024 represent the random sample taken

X=471 represent the people responded that they had used their cell phone while in a store within the last 30 days to call a friend or family member for advice about a purchase they were considering

$\hat p=\frac{471}{1024}=0.460$ estimated proportion of people responded that they had used their cell phone while in a store within the last 30 days to call a friend or family member for advice about a purchase they were considering

p= population proportion of people responded that they had used their cell phone while in a store within the last 30 days to call a friend or family member for advice about a purchase they were considering

Part a

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.