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Elden [556K]
3 years ago
15

Show work please :: I forgot how to solve

Mathematics
1 answer:
Vesna [10]3 years ago
4 0
40 pages = 80 minutes
210 pages = ? minutes
Well, take the first fact; 40 pages in 80 minutes and use it to find how many pages per minute.
40/80=1/2
1/2 pages = 1 minute
1 page = 2 minutes
210 pages = ? minutes
210*2=420
420 minutes = 7 hours
210 pages = 7 hours
Hope this helped :)
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Answer:

a) The 98% confidence interval would be given (0.182;0.218).

b) We are 98% confident that the true proportion of of England people who are deficient in Vitamin D is between 0.182 and 0.218.

c) If repeated samples were taken and the 98% confidence interval computed for each sample, 98% of the intervals would contain the population proportion.

d) Yes since the confidence interval not contains the value 0.25, we can refute the claim at 2% of significance.

Step-by-step explanation:

Data given and notation  

n=2700 represent the random sample taken    

X represent the people in England who are deficient in Vitamin D

\hat p=0.2 estimated proportion of England people who are deficient in Vitamin D

\alpha=0.02 represent the significance level

Confidence =0.98 or 98%

z would represent the statistic (variable of interest)    

p= population proportion of England people who are deficient in Vitamin D

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 98% confidence interval the value of \alpha=1-0.98=0.02 and \alpha/2=0.01, with that value we can find the quantile required for the interval in the normal standard distribution.

z_{\alpha/2}=2.33

And replacing into the confidence interval formula we got:

0.20 - 2.33 \sqrt{\frac{0.2(1-0.2)}{2700}}=0.182

0.20 + 2.33 \sqrt{\frac{0.2(1-0.2)}{2700}}=0.218

And the 98% confidence interval would be given (0.182;0.218).

Part b

We are 98% confident that the true proportion of of England people who are deficient in Vitamin D is between 0.182 and 0.218.

Part c

If repeated samples were taken and the 98% confidence interval computed for each sample, 98% of the intervals would contain the population proportion.

Part d

Yes since the confidence interval not contains the value 0.25, we can refute the claim at 2% of significance.

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