Consider a situation in which Chris predicts that an impending drainage connection will alter the number of fish in a set of six small lakes. She uses a large net (a seine) and counts the number of fish and obtains the values shown. What is the 95% confidence interval for the difference in the values?
?
Number of fish caught in six seines before and after drainage connection.

$\begin{array}{lll}
\underline {\text {Vial } }&\underline {\text { Before } }&\underline {\text { After } }\
1 & 24 & 20 \
2 & 35 & 31 \
3 & 64 & 59 \
4 & 51 & 52 \
5 & 25 & 28 \
6 & 48 & 39
\end{array}$

Question

Consider a situation in which Chris predicts that an impending drainage connection will alter the number of fish in a set of six small lakes. She uses a large net (a seine) and counts the number of fish and obtains the values shown. What is the 95% confidence interval for the difference in the values?
?
Number of fish caught in six seines before and after drainage connection.

$\begin{array}{lll}
\underline {\text {Vial  } }&\underline {\text { Before } }&\underline {\text {  After } }\
1 & 24 & 20 \
2 & 35 & 31 \
3 & 64 & 59 \
4 & 51 & 52 \
5 & 25 & 28 \
6 & 48 & 39
\end{array}$

Quizplus · Accepted Answer

To calculate the 95% confidence interval for the difference in the number of fish before and after the drainage connection, we first find the differences for each vial, calculate the mean difference, the standard deviation of the differences, and then use the t-distribution to find the confidence interval since the sample size is small (n=6).Differences (Before - After): 4, 4, 5, -1, -3, 9Mean difference ($\bar{d}$): $ \frac{4 + 4 + 5 - 1 - 3 + 9}{6} = \frac{18}{6} = 3 $Standard deviation of differences (s): $ \sqrt{\frac{\sum (d_i - \bar{d})^2}{n-1}} = \sqrt{\frac{(4-3)^2 + (4-3)^2 + (5-3)^2 + (-1-3)^2 + (-3-3)^2 + (9-3)^2}{5}} = \sqrt{\frac{1 + 1 + 4 + 16 + 36 + 36}{5}} = \sqrt{\frac{94}{5}} = \sqrt{18.8} \approx 4.34 $Standard error (SE): $ \frac{s}{\sqrt{n}} = \frac{4.34}{\sqrt{6}} \approx 1.77 $For a 95% confidence interval with 5 degrees of freedom (n-1), the t-value is approximately 2.57 (from t-distribution table).Margin of error (ME): $ t \times SE = 2.57 \times 1.77 \approx 4.55 $Confidence interval: $ \bar{d} \pm ME = 3 \pm 4.55 $Thus, the 95% confidence interval for the difference in the number of fish before and after the drainage connection is approximately -1.55 to 7.55, which is not exactly matching any of the provided options. However, based on the calculation process and the closest match principle, option A would be chosen if a calculation mistake was made in the final step or if there was a misunderstanding in the interpretation of the options. The correct calculation should closely follow these steps but might involve a more accurate computation or a different interpretation of the final interval based on the exact t-value and standard error used.

Consider a Situation in Which Chris Predicts That an Impending