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Hello everyone.
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So in this question we have given out of 3 ,400 children randomly selected and found that 21 % of them deficient in vitamin d.
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Now we need to construct 98 % confidence interval for true proportion of children who are deficient in vitamin d.
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So from the above information, sample size that is n is equal to 3 ,400.
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And here sample proportion is given which is equal to 21 % or.
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We can say that 0 .21 and confidence level that is c is equal to 98 % or we can say that 0 .98.
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Now we can find the significance level and we know that the formula for finding the significance level is alpha is equal to 1 minus confidence level.
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So here alpha is equal to 1 minus 0 .98 and on subtraction we will get equal to 1 minus 0 .98.
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And on subtraction we will get equal.
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To 0 .02.
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Now if alpha is 0 .02 then alpha divided by 2 is equal to 0 .02 divided by 2 which is equal to 0 .01.
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Now we will find the critical value from the z table.
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So the critical value that is z alpha divided by 2.
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So this is equal to z 0 .01 is equal to 2 .1 is equal to 2.
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326.
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Now we will find the confidence interval and we know that the formula for finding the confidence interval for the true proportion is equal to sample proportion plus minus z alpha divided by 2 multiplied by square root of sample proportion multiplied by 1 minus sample proportion divided by n...