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A survey of 85 families showed that 36 owned at least one DVD player. Find the 99% confidence interval estimate of the true proportion of families who own at least one DVD player. Place your limits, rounded to 3 decimal places, in the blanks. Place the lower limit in the first blank

User Parijat
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1 Answer

4 votes

Answer:
(0.367,\ 0.473)

Explanation:

The confidence interval for population mean is given by :-


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

, where
\hat{p} is the sample proportion, n is the sample size ,
z_(\alpha/2) is the critical z-value.

Given : Significance level :
\alpha:1-0.99=0.01

Sample size : n= 85

Critical value :
z_(\alpha/2)=2.576

Sample proportion:
\hat{p}=(36)/(85)\approx0.42

Now, the 99% confidence level will be :


\hat{p}\pmz_(\alpha/2)\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\\\\=0.42\pm(2.576)\sqrt{(0.42(1-0.42))/(85)}\\\\\approx0.42\pm0.053\\\\=(0.42-0.053,\ 0.42+0.053)=(0.367,\ 0.473)

Hence, the 99% confidence interval estimate of the true proportion of families who own at least one DVD player is
(0.367,\ 0.473)

User Kutzi
by
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