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denis-greek [22]

3 years ago

15

jackson has a rectangular piece of paper with a width of six units and a length of four units as shown below. He cuts the paper

into two equal right triangles, what is the area of each jackson's triangles

1 answer:

vfiekz [6]3 years ago

8 0

The are for the triangles is 3w X 2l = 6

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tigry1 [53]

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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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3 years ago