Question

in a survey, 10 people were asked how much they spent on their childs last birthday gift. the results were roughly bell - shaped with a mean of $34 and standard deviation of $5. construct a confidence interval at a 98% confidence level. give your answers to one decimal place. interpret your confidence interval in the context of this problem.

Explanation:

Step1: Identify the formula for confidence interval

For a sample with mean \(\bar{x}\), standard deviation \(s\), sample size \(n\), and confidence level \(1 - \alpha\), the confidence interval for the population mean (when the population standard deviation is unknown and the sample is from a normal population or \(n\) is large enough) is given by \(\bar{x}\pm t_{\alpha/2, n - 1}\times\frac{s}{\sqrt{n}}\). Here, \(\bar{x}=34\), \(s = 5\), \(n=10\), and the confidence level is \(98\%\), so \(\alpha=1 - 0.98 = 0.02\), \(\alpha/2=0.01\), and the degrees of freedom \(df=n - 1=9\).

Step2: Find the t - critical value

We need to find \(t_{0.01,9}\). Using a t - distribution table or a calculator, \(t_{0.01,9}\approx2.821\).

Step3: Calculate the margin of error (E)

The margin of error \(E=t_{\alpha/2, n - 1}\times\frac{s}{\sqrt{n}}\). Substitute the values: \(E = 2.821\times\frac{5}{\sqrt{10}}\). First, calculate \(\sqrt{10}\approx3.1623\), then \(\frac{5}{3.1623}\approx1.581\), and then \(E=2.821\times1.581\approx4.46\) (rounded to two decimal places for intermediate calculation, but we will keep more precision for the final answer).

Step4: Calculate the confidence interval

The confidence interval is \(\bar{x}\pm E\), so \(34\pm4.46\). But we need to round to one decimal place. First, calculate the lower and upper bounds:
Lower bound: \(34 - 4.46=29.54\approx29.5\) (wait, no, let's do the calculation more accurately). Wait, \(\frac{5}{\sqrt{10}}\approx1.5811\), \(t_{0.01,9} = 2.8214\) (more accurate value from t - table or calculator). Then \(E=2.8214\times1.5811\approx4.46\). So \(34\pm4.5\) (when rounded to one decimal place for the margin of error? Wait, no, let's recalculate:

\(\frac{5}{\sqrt{10}}\approx1.58113883\)

\(t_{0.01,9}=2.82143792\)

\(E = 2.82143792\times1.58113883\approx4.463\)

Rounding the margin of error to one decimal place: \(E\approx4.5\) (since the second decimal is 6, which is more than 5, we round up the first decimal: 4.463≈4.5).

And the mean is 34.0 (to one decimal place).

Answer:

\(34.0\pm4.5\)

For the interpretation: We are \(98\%\) confident that the true mean amount of money spent on a child's last birthday gift by all people (the population) is between \(34.0 - 4.5 = 29.5\) dollars and \(34.0+4.5 = 38.5\) dollars.