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Svetradugi [14.3K]
2 years ago
9

A researcher wishes to estimate the proportion of X-ray machines that malfunction. A random 275 sample of machines is taken, and

228 of the machines in the sample malfunction.
Based upon this, compute a 95% confidence interval for the proportion of all X-ray machines that malfunction. Then complete the table below.
Carry your intermediate computations to at least three decimal places. Round your answers to two decimal places.
What is the lower limit of the 95% confidence interval?
What is the upper limit of the 95% confidence interval?
Mathematics
1 answer:
Dominik [7]2 years ago
4 0

Answer:

Following are the solution to the given question:

Step-by-step explanation:

95\% Confidence Interval for both the percentage of all x-ray machines

p = the machinery's share is not working:

= \frac{228}{275}\\\\ = 0.829

\text{Margin of Error} = Z_{(\frac{\alpha}{2})} \times \sqrt{( p \times (1-p)}{n})

                         = 1.96 \times \sqrt{(0.829 \times \frac{0.171}{275})} \\\\= 1.96 \times 0.023 \\\\= 0.045

Lower 95\% Confidence interval = p - error margin = 0.829 - 0.045 = 0.784

Upper 95\% Confidence Interval = p + error margin= 0.829 + 0.045 = 0.874

So, 95\% Confidence Interval = ( 0.78 , 0.87 )

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<h3>Answer:  a = -3 and b = 5</h3>

=================================================

Work Shown:

Multiply top and bottom by \sqrt{2} to rationalize the denominator

\frac{10-\sqrt{18}}{\sqrt{2}}\\\\\frac{\sqrt{2}(10-\sqrt{18})}{\sqrt{2}*\sqrt{2}}\\\\\frac{\sqrt{2}*10-\sqrt{2}*\sqrt{18}}{\sqrt{2*2}}\\\\\frac{\sqrt{2}*10-\sqrt{2*18}}{\sqrt{4}}\\\\\frac{10\sqrt{2}-\sqrt{36}}{2}\\\\\frac{10\sqrt{2}-6}{2}\\\\\frac{-6+10\sqrt{2}}{2}\\\\\frac{2(-3+5\sqrt{2})}{2}\\\\-3+5\sqrt{2}\\\\

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