Question

suppose you use simple random sampling to select and measure 26 watermelons weights, and find they have a mean weight of 70 ounces. assume the population standard deviation is 7.6 ounces. based on this, construct a 99% confidence interval for the true population mean watermelon weight. give your answers as decimals, to two places: ± ounces question help: video post to forum

Explanation:

Step1: Find the z - score

For a 99% confidence interval, the significance level \(\alpha=1 - 0.99=0.01\). Then \(\alpha/2=0.005\). Looking up in the standard normal distribution table, the \(z\) - score \(z_{\alpha/2}=z_{0.005} = 2.576\)

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

The formula for the margin of error \(E\) when the population standard deviation \(\sigma\) is known is \(E = z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\)
Given \(n = 26\), \(\sigma=7.6\), and \(z_{\alpha/2}=2.576\)

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Step3: Calculate the confidence interval

The confidence interval for the population mean \(\mu\) is \(\bar{x}\pm E\)
Given \(\bar{x} = 70\)
The confidence interval is \(70\pm3.84\)

Answer:

\(70\pm3.84\)