Question

high school graduation gifts a random survey of 24 couples who had children graduating from high school found that they spent an average of $128.33 on graduation gifts. the standard deviation of the sample is $5.70. find the 95% confidence interval for the mean amount of money couples spent for graduation gifts. round intermediate calculations to three decimal places, and final answers to two decimal places. $\square<\mu<$\square

Explanation:

Step1: Determine the critical value

Since the sample size \(n = 24\) (small - sample, \(n<30\)) and we want a \(95\%\) confidence interval, the degree of freedom \(df=n - 1=24 - 1 = 23\). For a \(95\%\) confidence interval (\(\alpha=1 - 0.95 = 0.05\), \(\alpha/2=0.025\)), using the \(t\) - distribution table or a calculator, the critical value \(t_{\alpha/2}\) is \(t_{0.025,23}=2.069\).

Step2: Calculate the margin of error

The formula for the margin of error \(E\) for a \(t\) - distribution confidence interval for the mean is \(E = t_{\alpha/2}\frac{s}{\sqrt{n}}\), where \(s = 5.70\) (sample standard deviation) and \(n = 24\).

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Step3: Calculate the confidence interval

The formula for the confidence interval for the population mean \(\mu\) (when \(\sigma\) is unknown) is \(\bar{x}-E<\mu<\bar{x} + E\), where \(\bar{x}=128.33\) (sample mean).

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Answer:

\(125.92<\mu<130.74\)