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Gnesinka [82]
2 years ago
12

Help asap pleaseeeee

Mathematics
2 answers:
Ludmilka [50]2 years ago
8 0

Answer: 540sqrt2

Step-by-step explanation:

nydimaria [60]2 years ago
8 0
540SQ2










Hope it helps


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Use the normal distribution to find a confidence interval for a proportion p given the relevant sample results. Give the best po
mina [271]

Answer:

(a) The point estimate for the population proportion <em>p</em> is 0.34.

(b) The margin of error for the 99% confidence interval of population proportion <em>p</em> is 0.055.

(c) The 99% confidence interval of population proportion <em>p</em> is (0.285, 0.395).

Step-by-step explanation:

A point estimate of a parameter (population) is a distinct value used for the estimation the parameter (population). For instance, the sample mean \bar x is a point estimate of the population mean <em>μ</em>.

Similarly, the the point estimate of the population proportion of a characteristic, <em>p</em> is the sample proportion \hat p.

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

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

The margin of error for this interval is:

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

The information provided is:

\hat p=0.34\\n=500\\(1-\alpha)\%=99\%

(a)

Compute the point estimate for the population proportion <em>p</em> as follows:

Point estimate of <em>p</em> = \hat p = 0.34

Thus, the point estimate for the population proportion <em>p</em> is 0.34.

(b)

The critical value of <em>z</em> for 99% confidence level is:

z={\alpha/2}=z_{0.01/2}=z_{0.005}=2.58

*Use a <em>z</em>-table for the value.

Compute the margin of error for the 99% confidence interval of population proportion <em>p</em> as follows:

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

          =2.58\sqrt{\frac{0.34(1-0.34)}{500}}

          =2.58\times 0.0212\\=0.055

Thus, the margin of error for the 99% confidence interval of population proportion <em>p</em> is 0.055.

(c)

Compute the 99% confidence interval of population proportion <em>p</em> as follows:

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

CI=\hat p\pm MOE

     =0.34\pm 0.055\\=(0.285, 0.395)

Thus, the 99% confidence interval of population proportion <em>p</em> is (0.285, 0.395).

6 0
3 years ago
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