A study of 4,430 U.S. adults found that 2,913 were obese or overweight. The 95% confidence interval for the proportion of adults in U.S. who

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A study of 4,430 U.S. adults found that 2,913 were obese or overweight. The 95% confidence interval for the proportion of adults in U.S. who are obese or overweight is

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Hadley 1 hour 2021-09-15T23:52:55+00:00 1 Answer 0

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    2021-09-15T23:54:46+00:00

    Answer: (0.644,0.672)

    Step-by-step explanation:

    Let p be the proportion of the adults in U.S. who are obese or overweight.

    Confidence interval for p will be :

    \hat{p}\pm z\sqrt{\frac{\hat{p}(1-\hat{p)}}{n}}, where \hat{p} is sample proportion , n is sample size , and z is the critical z value as per confidence level.

    As per given, we have

    n= 4430

    \hat{p}=\frac{2913}{4430}\approx0.658

    Critical z-value for 95% confidence level is 1.96.

    Then, the 95% confidence interval for the proportion of adults in U.S. who are obese or overweight  would be:

    0.658\pm (1.96)\sqrt{\frac{0.658(1-0.658)}{4430}}\\\\=0.658\pm1.96\times0.0071272851865\\\\=0.658\pm0.0139694789655\\\\\approx(0.658-0.014,0.658+0.014)\\\\=(0.644,0.672)

    Hence, the required 95% confidence interval would be (0.644,0.672).

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