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A study was commissioned to find the mean weight of the residents in certain town.The study examined a random sample of 37 residents and found the mean weight tobe 158 pounds with a standard deviation of 27 pounds. Determine a 95% confidenceinterval for the mean, rounding all values to the nearest tenth.

User Immu
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We know that the sample n= 37, the mean is 158 and standard deviation is 27. At 95% confidence the interval is given by


\bar{x}\pm z*\frac{\sigma}{\sqrt[]{n}}

where z is equal to 1.96. By substituting the given values into this formula, we get


158\pm(1.96)\frac{27}{\sqrt[]{37}}

Then, the interval is


158\pm8.699

Finally, by rounding up to the nearest tenth, the interval is


\begin{gathered} 158\pm8.7 \\ or \\ 149.3\le158\le166.7 \end{gathered}

The value for z at 95% confidence is given in the followin table:

A study was commissioned to find the mean weight of the residents in certain town-example-1
User Achennu
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