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overweight men for a random sample of 60 overweight men, the mean of th…

Question

overweight men for a random sample of 60 overweight men, the mean of the number of pounds that they were overweight was 32. the standard deviation of the population is 4.2 pounds.
(a) the best point estimate of the mean is 32 pounds.
(b) find the 95% confidence interval of the mean of these pounds. round intermediate answers to at least three decimal places. round your final answers to one decimal place.
\\( 1 < \mu < \\)

Explanation:

Part (b)

Step1: Identify the formula for confidence interval

For a confidence interval for the population mean when the population standard deviation (\(\sigma\)) is known, the formula is:
\(\bar{x} \pm z_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}}\)
where:

  • \(\bar{x} = 32\) (sample mean),
  • \(\sigma = 4.2\) (population standard deviation),
  • \(n = 60\) (sample size),
  • \(z_{\alpha/2}\) is the z - score for a 95% confidence level. For a 95% confidence level, \(\alpha=1 - 0.95 = 0.05\), so \(\alpha/2=0.025\). The \(z\) - score \(z_{0.025}\) is approximately 1.96.

Step2: Calculate the margin of error (\(E\))

The margin of error \(E=z_{\alpha/2}\cdot\frac{\sigma}{\sqrt{n}}\)
Substitute the values:
\(E = 1.96\times\frac{4.2}{\sqrt{60}}\)
First, calculate \(\sqrt{60}\approx7.746\)
Then, \(\frac{4.2}{7.746}\approx0.542\)
Then, \(E = 1.96\times0.542\approx1.062\) (rounded to three decimal places)

Step3: Calculate the confidence interval

The lower limit of the confidence interval is \(\bar{x}-E\) and the upper limit is \(\bar{x} + E\)
Lower limit: \(32-1.062 = 30.938\approx30.9\) (rounded to one decimal place)
Upper limit: \(32 + 1.062=33.062\approx33.1\) (rounded to one decimal place)

Answer:

The 95% confidence interval is \( 30.9 < \mu < 33.1 \)