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Lilit [14]
3 years ago
5

Margin of error and the confidence interval. A study of stress on the campus of your university reported a mean stress level of

78 (on a 0 to 100 scale with a higher score indicating more stress) with a margin of error of 5 for 95% confidence. The study was based on a random sample of 64 undergraduates. (a) Give the 95% confidence interval. (b) If you wanted 99% confidence for the same study, would your margin of error be greater than, equal to, or less than 5
Mathematics
1 answer:
victus00 [196]3 years ago
3 0

Answer:

(a) The 95% confidence interval for the population mean stress level is (73, 83).

(b) Increasing the confidence level to 99% from 95% the margin of error would be greater than 5.

Step-by-step explanation:

The (1 - <em>α</em>) % confidence interval for population mean is:

CI=\bar x\pm MOE

The information provided is:

\bar x = 78

Confidence level = 95%

MOE = 5

(a)

Compute the 95% confidence interval for the population mean stress level as follows:

CI=\bar x\pm MOE\\=78\pm5\\=(73, 83)

Thus, the 95% confidence interval for the population mean stress level is (73, 83).

(b)

The formula to compute the margin of error (MOE) is:

MOE=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}

The margin of error is affected by:

  1. Standard deviation
  2. Sample size
  3. Confidence level.

On increasing the confidence level the critical value of <em>z</em> increases.

z_{90\%}=1.645\\z_{95\%}=1.96\\z_{99\%}=2.58

And if the critical value is increased then the margin of error will also increase.

Thus, increasing the confidence level to 99% from 95% the margin of error would be greater than 5.

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If 0\le x, then \cos x=\dfrac12 is true for x=\dfrac\pi3 and x=\dfrac{5\pi}3. Then to account for all other possible values, we add a multiple of 2\pi, so that x=\dfrac\pi3+2n\pi or x=\dfrac{5\pi}3+2n\pi for integers n.

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So, the sum of all the three digits multiples of 4 is

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We know how to sum the first N integers, but we need to sum the first 249 integers starting from 25, so we can rewrite our sum using this little trick:

\displaystyle \sum_{k=25}^{249}k = \sum_{k=1}^{249}k - \sum_{k=1}^{24}k

In other words, we're summing all the first 249 integers, but then we remove the first 24. As a result, we have the sum of all integers from 25 to 249:

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