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nordsb [41]
2 years ago
11

In a​ poll, adults in a region were asked about their online vs.​ in-store clothes shopping. One finding was that ​% of responde

nts never​ clothes-shop online. Find and interpret a ​% confidence interval for the proportion of all adults in the region who never​ clothes-shop online.
Mathematics
1 answer:
White raven [17]2 years ago
7 0

The question is incomplete. The complete question is :

In a​ poll, 1100 adults in a region were asked about their online vs.​ in-store clothes shopping. One finding was that 43​% of respondents never​ clothes-shop online. Find and interpret a ​95% confidence interval for the proportion of all adults in the region who never​ clothes-shop online.

Solution :

95% confidence interval for p is :

$\hat p - Z_{\alpha/2}\times \sqrt{\frac{\hat p(1-\hat p)}{n}} < p < \hat p + Z_{\alpha/2}\times \sqrt{\frac{\hat p(1-\hat p)}{n}}$

$0.43 - 1.96\times \sqrt{\frac{0.43(1-0.43)}{1100}} < p < 0.43 + 1.96\times \sqrt{\frac{0.43(1-0.43)}{1100}}$

0.401 < p < 0.459

Therefore, 95% confidence interval is from 0.401 to 0.459

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Answer:

$6 = cost of small box

$8 = cost of large box

Step-by-step explanation:

Let s = cost of small box

     l  = cost of large box

(1)    12s + 3l = 96            (2)     6s + 6l = 84

    Multiply by -2                    <u> -24s - 6l = -192</u>

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