In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the z-score that has a p-value of $1 - \frac{\alpha}{2}$ .

They randomly survey 387 drivers and find that 298 claim to always buckle up.

This means that $n = 387, \pi = \frac{298}{387} = 0.77$

84% confidence level

So $\alpha = 0.16$ , z is the value of Z that has a p-value of $1 - \frac{0.16}{2} = 0.92$ , so $Z = 1.405$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.77 - 1.405\sqrt{\frac{0.77*0.23}{387}} = 0.74$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.77 + 1.405\sqrt{\frac{0.77*0.23}{387}} = 0.8$

The 84% confidence interval for the population proportion that claim to always buckle up is (0.74, 0.80).

6 0

3 years ago

Please help me idk this

crimeas [40]

Answer:

Step-by-step explanation:

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6 0

3 years ago

In ΔUVW, the measure of ∠W=90°, WV = 24, UW = 7, and VU = 25. What is the value of the sine of ∠U to the nearest hundredth?

Alika [10]

Answer:

sinU = 0.96

Step-by-step explanation:

sinU = $\frac{opposite}{hypotenuse}$ = $\frac{WV}{UV}$ = $\frac{24}{25}$ = 0.96

7 0

3 years ago