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

Answer:

$9.9676 - 2.326*0.5904 =8.594$

$9.9676 + 2.326*0.5904 =11.341$

Step-by-step 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:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$   (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:

$\bar X = \frac{8.8104 +11.1248}{2}= 9.9676$

And we can find the margin of error with:

$ME= \frac{11.1248- 8.8104}{2}= 1.1572$

The margin of error for this case is given by:

$ME = t_{\alpha/2}\frac{s}{\sqrt{n}} = t_{\alpha/2} SE$

And we can solve for the standard error:

$SE = \frac{ME}{t_{\alpha/2}}$

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

$SE = \frac{1.1572}{1.96}= 0.5904$

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:

$9.9676 - 2.326*0.5904 =8.594$

$9.9676 + 2.326*0.5904 =11.341$