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julia-pushkina
16 days ago
13

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 [8.8K]16 days ago
8 0

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

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