## Related questions with answers

Calculating a $95 \%$ confidence interval $(95 \% \mathrm{Cl})$ provides a good estimate of how close the sample mean is to the true mean. The mean $\pm$ the $95 \%$ $\mathrm{Cl}$ gives the $95 \%$ confidence limits. On average, 95 times out of 100 , the true population mean will lie within the confidence limits. The $95 \% \mathrm{Cl}$ can be plotted onto graphs to determine significance. If they overlap there is no significant difference between the data. The $95 \% \mathrm{Cl}$ for student $\mathrm{A}$ is $1.27$ and for student $\mathrm{B}$ is $2.37$ at $\mathrm{p}=0.05$.

(a) Plot the mean dispensing volume and $95 \% \mathrm{Cl}$ for both sets of data on the grid right). Plot as a column graph with error bars.

(b) Is the difference significant? Why or why not?

Solution

VerifiedConsider the information provided in the exercise and that the 95% CI for student A is 1.27 and for student B is 2.37 and the p-value is 0.05. Then:

(a) plot the mean dispensing volume and 95% CI for both students.

(b) explain if the difference is significant and why

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