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Debora [2.8K]
4 years ago
6

A random sample of medical files is used to estimate the proportion p of all people who have blood type b.

Mathematics
1 answer:
makvit [3.9K]4 years ago
7 0

Answer:

a) 106

b) 53

Step-by-step explanation:

Minimum required medical files can be found using the formula

n≥ p×(1-p) × (\frac{z}{ME} )^2 where

  • n is the minimum required medical files
  • p is the estimated proportion for blood type b
  • z is the corresponding z-score for two tailed 85% confidence level
  • ME is the margin of error in the estimation

For the case (a)

If there is no preliminary estimate for p, then it is estimated as p=0.5

corresponding z-score for two tailed 85% confidence level is 1.44

and ME is 0.07.

Putting all in the formula;

n≥ 0.5×0.5 × (\frac{1.44}{0.07} )^2  ≈ 106

For the case (b)

The only difference is p=\frac{13}{90}≈0.144

n≥ 0.144×0.856 × (\frac{1.44}{0.07} )^2  ≈ 53

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95% confidence interval for the difference between the proportions of males and females who have the blood disorder is [-0.064 , 0.014].

Step-by-step explanation:

We are given that a certain geneticist is interested in the proportion of males and females in the population who have a minor blood disorder.

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<em>Here for constructing 95% confidence interval we have used Two-sample z proportion statistics.</em>

<u>So, 95% confidence interval for the difference between the population proportions, </u><u>(</u>p_1-p_2<u>)</u><u> is ;</u>

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P(-1.96 < \frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} } < 1.96) = 0.95

P( -1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} } < {(\hat p_1-\hat p_2)-(p_1-p_2)} < 1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} } ) = 0.95

P( (\hat p_1-\hat p_2)-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} } < (p_1-p_2) < (\hat p_1-\hat p_2)+1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} } ) = 0.95

<u>95% confidence interval for</u> (p_1-p_2) =

[(\hat p_1-\hat p_2)-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }, (\hat p_1-\hat p_2)+1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }]

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