Question

An experiment was conducted to record the jumping distances of paper frogs made from construction paper. Based on the sample, the corresponding 95% confidence interval for the mean jumping distance is (8.8104, 11.1248)cm. What is the corresponding 98% confidence interval for the mean jumping distance?

Answer 1

`9.9676 - 2.326*0.5904 =8.594`

`9.9676 + 2.326*0.5904 =11.341`

Explanation:

Notation

`\bar X` represent the sample mean

`\mu` population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size

Solution to the problem

The confidence interval for the mean is given by the following formula:

`\bar X \pm t_(\alpha/2)(s)/(√(n))` (1)

For this case the 9% confidence interval is given by:

`8.8104 \leq \mu \leq 11.1248`

We can calculate the mean with the following:

`\bar X = (8.8104 +11.1248)/(2)= 9.9676`

And we can find the margin of error with:

`ME= (11.1248- 8.8104)/(2)= 1.1572`

The margin of error for this case is given by:

`ME = t_(\alpha/2)(s)/(√(n)) = t_(\alpha/2) SE`

And we can solve for the standard error:

`SE = (ME)/(t_(\alpha/2))`

The critical value for 95% confidence using the normal standard distribution is approximately 1.96 and replacing we got:

`SE = (1.1572)/(1.96)= 0.5904`

Now for the 98% confidence interval the significance is
`\alpha=1-0.98= 0.02` and
`\alpha/2 = 0.01` the critical value would be 2.326 and then the confidence interval would be:

`9.9676 - 2.326*0.5904 =8.594`

`9.9676 + 2.326*0.5904 =11.341`