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Ten students at a local college were randomly selected and asked how many hours they spend studying on the weekend. The data collected is shown in the table below.
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|}
\hline Student & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\
\hline \begin{tabular}{c}
Hours Spent \\
Studying
\end{tabular} & 4 & 6 & 1 & 0 & 5 & 8 & 2 & 3 & 3 & 4 \\
\hline
\end{tabular}

The formulas below are provided for reference, where [tex]$E$[/tex] is the margin of error, [tex]$\sigma$[/tex] is the population standard deviation, and [tex]$n$[/tex] is the sample size.
\begin{tabular}{|c|c|}
\hline \begin{tabular}{c}
Margin of Error \\
(95\% Confidence)
\end{tabular} & \begin{tabular}{c}
Minimum Sample Size Needed \\
(95\% Confidence)
\end{tabular} \\
\hline[tex]$E=1.96 \cdot \frac{\sigma}{\sqrt{n}}$[/tex] & [tex]$n=\left(\frac{1.96 \cdot \sigma}{E}\right)^2$[/tex] \\
\hline
\end{tabular}

Assuming the data for the population is normally distributed with a standard deviation of 2.4 hours, which of the statements are true?

- If the desired margin of error is 2 hours, then the sample mean ensures with [tex]$95\%$[/tex] confidence that the sample size is large enough.
- It is a good decision for the school newspaper to use this study and report that the...



Answer :

GinnyAnswer
Statements that can be confirmed as true based on the given data and the derived values are:

1. The sample mean of the hours spent studying by these ten students is 3.6 hours.
- This is determined by calculating the average of the data values: [tex]\( \frac{4 + 6 + 1 + 0 + 5 + 8 + 2 + 3 + 3 + 4}{10} = 3.6 \)[/tex].

2. The margin of error with the provided sample size (10) and 95% confidence level is approximately 1.4875 hours.
- The margin of error [tex]\( E \)[/tex] is calculated using the formula [tex]\( E = 1.96 \cdot \frac{\sigma}{\sqrt{n}} \)[/tex]. Substituting [tex]\(\sigma = 2.4\)[/tex] and [tex]\(n = 10\)[/tex]: [tex]\( E = 1.96 \cdot \frac{2.4}{\sqrt{10}} \approx 1.4875 \)[/tex].

3. The minimum sample size needed to achieve a desired margin of error of 2 hours with 95% confidence is approximately 5.532.
- The required sample size [tex]\( n \)[/tex] is calculated using the formula [tex]\( n = \left(\frac{1.96 \cdot \sigma}{E}\right)^2 \)[/tex]. Substituting [tex]\(\sigma = 2.4\)[/tex] and [tex]\( E = 2 \)[/tex]: [tex]\( n = \left(\frac{1.96 \cdot 2.4}{2}\right)^2 \approx 5.532 \)[/tex].

Hence, the statements that correctly relate to the above computations are:

- The sample mean of the hours spent studying by these ten students is 3.6 hours.
- The margin of error with the provided sample size of 10 and a 95% confidence level is approximately 1.4875 hours.
- The minimum sample size needed to achieve a desired margin of 2 hours with 95% confidence is approximately 5.532, which suggests that the current sample size of 10 is sufficient.

The statement about the sample size ensuring that the desired margin of error is achieved with 95% confidence is true given the current sample size of 10 is more than the calculated minimum sample size of approximately 5.532.