Question

you measure 50 turtles weights, and find they have a mean weight of 47 ounces. assume the population standard deviation is 3.2 ounces. based on this, construct a 90% confidence interval for the true population mean turtle weight. give your answers as decimals, to two places; however use a critical value rounded to four decimal places. ± ounces

Explanation:

Step1: Find the critical value

For a 90% confidence interval, the significance level \(\alpha = 1 - 0.90=0.10\), and \(\alpha/2=0.05\).
Using the standard normal distribution \(Z\)-table or a calculator, the critical value \(z_{\alpha/2}\) is \(z_{0.05}\approx1.6449\) (rounded to four decimal places).

Step2: Calculate the margin of error

The formula for the margin of error \(E\) when the population standard deviation \(\sigma\) is known is \(E = z_{\alpha/2}\times\frac{\sigma}{\sqrt{n}}\).
Given \(n = 50\), \(\sigma=3.2\), and \(z_{\alpha/2}=1.6449\).

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Step3: Find the confidence interval

The confidence interval is \(\bar{x}\pm E\), where \(\bar{x} = 47\).

Answer:

\(47\pm0.74\) ounces