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strojnjashka [21]
4 years ago
9

Faculty members at Lowell Place High School want to determine whether there are enough students to have a Valentine's Day Formal

. Eighty-eight of the 200 students said they would attend the Valentine's Day Formal. Construct and interpret a 90% confidence interval for p.
Mathematics
1 answer:
Fittoniya [83]4 years ago
5 0

Answer:

The 90% confidence interval is (0.383,0.497)    

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 200

Number of children that would attend  Valentine's Day Forma, x = 88

\hat{p} = \dfrac{x}{n} = \dfrac{88}{200} = 0.44

90% Confidence interval:

\hat{p}\pm z_{stat}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}

z_{critical}\text{ at}~\alpha_{0.10} = \pm 1.64

Putting the values, we get:

0.44\pm 1.64(\sqrt{\dfrac{0.44(1-0.44)}{200}}) = 0.44\pm 0.057\\\\=(0.383,0.497)

Interpretation:

The 90% confidence interval is (0.383,0.497). We are 90% confident that the proportion of children attending Valentine's Day Formal is between 38.3% and 49.7%

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Step-by-step explanation:

Data provided in the question:

Mean = $19,800

Standard deviation, s = $2,900

Now,

z score = [ X - Mean ] ÷ s

a) The procedure costs between $18,000 and $22,000

For X = $22,000

z-score = [ $22,000 - $19,800 ] ÷ $2,900

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For X = $18,000

z-score = [ $18,000 - $19,800 ] ÷ $2,900

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The procedure costs more than $17,250.

P (z > -0.87931  ) = 0.8103834

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