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Rzqust [24]
3 years ago
13

Please help out with this question..

Mathematics
1 answer:
Sophie [7]3 years ago
8 0

Answer:

Step-by-step explanation:

C = 16 * 3 = 48

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A survey of 85 families showed that 36 owned at least one DVD player. Find the 99% confidence interval estimate of the true prop
attashe74 [19]

Answer:

0.424 - 2.58\sqrt{\frac{0.424(1-0.424)}{85}}=0.286

0.424 + 2.58\sqrt{\frac{0.424(1-0.424)}{85}}=0.562

The 99% confidence interval would be given by (0.286;0.562)

Step-by-step explanation:

Information given:

X= 36 represent the families owned at least one DVD player

n= 85 represent the total number of families

\hat p=\frac{36}{85}= 0.424 represent the estimated proportion of families owned at least one DVD player

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 99% of confidence, our significance level would be given by \alpha=1-0.99=0.01 and \alpha/2 =0.05. And the critical value would be given by:

z_{\alpha/2}=-2.58, z_{1-\alpha/2}=2.58

The confidence interval for the mean is given by the following formula:  

\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}

If we replace the values obtained we got:

0.424 - 2.58\sqrt{\frac{0.424(1-0.424)}{85}}=0.286

0.424 + 2.58\sqrt{\frac{0.424(1-0.424)}{85}}=0.562

The 99% confidence interval would be given by (0.286;0.562)

8 0
3 years ago
On the first day of school bethanys family donated $60 to the field trip fundraiser. what percent of their goal did bethanys fam
Romashka-Z-Leto [24]

Answer:

B

Step-by-step explanation:

Got it right on edge, plz mark brainliest

7 0
3 years ago
Read 2 more answers
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