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Fiesta28 [93]
2 years ago
8

PLEASE HELP ME WITH THIS QUESTION ^^

Mathematics
1 answer:
AURORKA [14]2 years ago
4 0
I’m sorry I don’t get that whole situation
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The pre–image, A, was dilated about the origin. It was then transformed in another way to get A'.
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Step-by-step explanation:

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In a survey of 603 adults, 98 said that they regularly lie to people conducting surveys. Create a 99% confidence interval for th
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Answer:

The population proportion is estimated to be with 99% confidence within the interval (0.1238, 0.2012).

Step-by-step explanation:

The formula for estimating the population proportion by a confidence interval is given by:

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

Where:

\hat{p} is the sample's proportion of success, which in this case is the people that regularly lie during surveys,

z_{\alpha /2} is the critical value needed to find the tails of distribution related to the confidence level,

n is the sample's size.

<u>First</u> we compute the \hat{p} value:

\hat{p}=\frac{successes}{n}=\frac{98}{603}=0.1625

<u>Next</u> we find the z-score at any z-distribution table or app (in this case i've used StatKey):

z_{\alpha /2}=2.576

Now we can replace in the formula with the obtained values to compute the confidence interval:

\hat{p}\pm z_{\alpha /2}\times\sqrt{\frac{\hat{p}\times(1-\hat{p})}{n}}=0.1625\pm 2.576\times\sqrt{\frac{0.1625\times(1-0.1625)}{603}}=(0.1238, 0.2012)

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