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A study was commissioned to find the mean weight of the residents in a certain town. The study examined a random sample of 132 residents and found the mean weight to be 162 pounds with a standard deviation of 35 pounds. At the 95% confidence level, find the margin of error for the mean, rounding to the nearest tenth

Sagot :

Answer: 6.1

We are 95% confident that the true mean weight of the population is within 6.1 pounds of the sample mean 162.

At the 95% confidence interval, the margin of error for the mean is  6.1 pounds

What is margin error?

The margin of error is "a statistic expressing the amount  of random sampling error in the results of a survey".

According to the question,

Number of random samples (n) = 132 residents

Mean of weight of random samples (x⁻)= 162

Standard deviation of random samples (σ)= 35 pounds

In order to find at the 95% confidence interval, the margin of error for the mean.

Margin error =  z (σ/√n)

The critical region 95% confidence interval under normal curve is 1.96.

= 1.96 (35/√132)

=1.96(35/11.48)

=1.96(3.0488)

= 5.9708636  ≈ 6.1 pounds

Hence, at the 95% confidence interval, the margin of error for the mean is  6.1 pounds.

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