An experiment was conducted to record the jumping distances of paper frogs made from construction paper. Based on the sample, th

Question

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An experiment was conducted to record the jumping distances of paper frogs made from construction paper. Based on the sample, th

e 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?

Mathematics

1 answer:

tester [12.3K] · Answer 1 · 2026-03-12T00:41:45+03:00

Answer:

$9.9676 - 2.326*0.5904 =8.594$

$9.9676 + 2.326*0.5904 =11.341$

Step-by-step explanation:

Notation

$\bar X$ is the sample mean

$\mu$ indicates the population mean (the variable of interest)

s signifies the sample standard deviation

n denotes the sample size

Solution to the problem

The mean's confidence interval is derived from the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In this instance, the 9% confidence interval corresponds to:

$8.8104 \leq \mu \leq 11.1248$

We can determine the mean using the following:

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

Additionally, the margin of error can be calculated as:

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

The margin of error for this situation is expressed as:

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

Next, we find the standard error:

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

The critical value for a 95% confidence interval using a normal standard distribution is roughly 1.96, and substituting gives us:

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

For the 98% confidence interval, the significance corresponds to $\alpha=1-0.98= 0.02$ and $\alpha/2 = 0.01$ with a critical value of 2.326, yielding a confidence interval of:

$9.9676 - 2.326*0.5904 =8.594$

$9.9676 + 2.326*0.5904 =11.341$