## Related questions with answers

Suppose a sample of $11$ paired differences that has been randomly selected from a normally distributed population of paired differences yields a sample mean of $103.5$ and a sample standard deviation of $5.$

Calculate $95$ percent and $99$ percent confidence intervals for $\mu_d=\mu_1-\mu_2$.

Solutions

VerifiedThe goal of the exercise is to calculate 95 percent and 99 percent confidence intervals for

$\mu_{ d}=\mu _{1} - \mu_{2}$

Then we should check that the difference between $\mu _{1}$ and $\mu _{2}$ exceeds 100?

Let us prepare our given information as follows:

The number of the sample is 11 ( normally distributed),

The sample mean is $\bar{d}=103.5$;

The sample standard deviation is $s_{d}=5$.

It is required of us to determine the interval for the mean of a population of paired differences with a certain percent of confidence. In this case we have to use this formula which will help us to determine a $100(1-\alpha)\%$ confidence interval for the mean of a population of paired differences and that formula is:

$\left[ \=d \pm t_{\alpha/2}\frac{s_d}{\sqrt{n}} \right].$

Here $\=d$ represents the value of the difference between two sample means, and $s_d$ represents the value of a sample standard deviation. The value of $n$ will tell us how large the sample size is. Also, what is important to know is that the value of $t_{\alpha/2}$ depends on $(n-1)$ degrees of freedom.

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