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An airline transports over 10,000 passengers daily, and the airline is curious what proportion of their passengers use mobile boarding passes instead of paper passes. They take an SRS of 80 passengers and find that 60 of them use mobile boarding passes. Based on this sample, which of the following is a 95% confidence interval for the proportion of passengers who use mobile boarding passes?A. (0.625, 0.875) B. (0.637, 0.863) C. (0.655, 0.845) D. (0.670,0.830)

Respuesta :

quentinmichael

Answer:

(0.655,0.845)

Step-by-step explanation:

p

^

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0.75

0.75

0.75

0.75

0.75

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±z  

∗

 

n

p

^

​  

(1−  

p

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)

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±(1.96)  

80

0.75(1−0.75)

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±(1.96)(0.048412)

±0.094888;

−0.094888=0.655112

+0.094888=0.844888

​  

Following are the calculation to the z value:

Given:

[tex](n) = 80\\\\ (X) = 60\\\\ C.I= 95\%\\[/tex]

To find:

z value=?

Solution:

[tex](n) = 80\\\\ (X) = 60\\\\ (\hat{p}) = \frac{X}{n} = \frac{60}{80} = 0.75\\\\ [/tex]

In this case, we must compute a [tex]95\%[/tex] confidence interval for the fraction of passengers who use mobile boarding cards, which may be calculated as

[tex]\to \hat{p} \pm Z_{\frac{\alpha}{2}} \times \sqrt{(\frac{\hat{p}\times (1-\hat{p})}{n})}\\\\ \to C.I = 0.95\\\\ [/tex]

The level of importance [tex] (\alpha) = 1 - 0.95 = 0.05[/tex]

[tex]\to \frac{\alpha}{2} = \frac{0.05}{2} = 0.025[/tex]

from the Z table, we found[tex] Z_{\frac{\alpha}{2}} = 1.96[/tex] So [tex]95\%[/tex] confidence interval is

[tex]\to 0.75 \pm 1.96 \times \sqrt{(\frac{0.75\times (1-0.75)}{80})}\\\\ \to 0.75 \pm 1.96 \times \sqrt{(\frac{0.75\times 0.25}{80})}\\\\ \to 0.75 \pm 1.96 \times \sqrt{(\frac{0.1875}{80})}\\\\ \to 0.75 \pm 1.96 \times \sqrt{(0.00234375)}\\\\ \to 0.75 \pm 1.96 \times 0.05\\\\ \to 0.75 \pm 0.098\\\\ \to 0.652

So the [tex]95\%[/tex] confidence interval for the proportion of passengers that utilize mobile boarding cards is somewhere between 0.652 and 0.848.

Therefore, the answer is "Option C".

Learn more:

brainly.com/question/22867285

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