A botanist measures the heights of 16 seedlings and obtains a mean and standard deviation of 72.5 cm and 4.5 cm,respectively.Find the 95 percent confidence interval for the mean height of seedlings in the population from which the sample was selected.
A) $\left[ \begin{array} { l l }
62.91 & 82.09
\end{array} \right]$
B) $[ 70.30$$ \quad$$ \quad$$74.70 ]$
C) $[ 70.10$$ \quad$$ \quad$$74.90 ]$
D) $[ 70.54$$ \quad$$ \quad$$74.46 ]$
E) $[ 71.90$$ \quad$$ \quad$$73.10 ]$

Question

A botanist measures the heights of 16 seedlings and obtains a mean and standard deviation of 72.5 cm and 4.5 cm,respectively.Find the 95 percent confidence interval for the mean height of seedlings in the population from which the sample was selected.
A) $\left[ \begin{array} { l l } 
62.91 & 82.09
\end{array} \right]$
B) $[ 70.30$$ \quad$$ \quad$$74.70 ]$
C) $[ 70.10$$ \quad$$ \quad$$74.90 ]$
D) $[ 70.54$$ \quad$$ \quad$$74.46 ]$
E) $[ 71.90$$ \quad$$ \quad$$73.10 ]$

Quizplus · Accepted Answer

The 95% confidence interval is calculated using the formula: mean ± (t-value * (standard deviation/√n)). For 15 degrees of freedom, the t-value is approximately 2.131. Thus, the interval is 72.5 ± (2.131 * (4.5/√16)) = [70.10, 74.90].

A Botanist Measures the Heights of 16 Seedlings and Obtains [62.9182.09]\left[ \begin{array} { l l } 62.91 & 82.09\end{array} \right][62.91​82.09​]

A Botanist Measures the Heights of 16 Seedlings and Obtains $\left[ \begin{array} { l l } 62.91 & 82.09\end{array} \right]$