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?

Step-by-step explanation:

Notation

represent the sample mean

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:

  (1)

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

We can calculate the mean with the following:

And we can find the margin of error with:

The margin of error for this case is given by:

And we can solve for the standard error:

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

Now for the 98% confidence interval the significance is and the critical value would be 2.326 and then the confidence interval would be:

1: -2 3/4, -2 1/2, -14, 0, 3, 8 1/3

2: -4, -3, -2 1/2, -2 1/4, 3 1/4,

step-by-step explanation: