Question

Determine the 95% confidence interval for the difference of the sample means. then complete the

Answer 1

Using the z-distribution, it is found that the 95% confidence interval is of -1.38 to 1.38, and 1.74 is outside the interval.

What is a confidence interval for the difference in sample means?

It is given by:

$\overline{x} \pm zs$

In which:

• $\overline{x}$ is the estimate of the difference.

• s is the standard error.

• z is the critical value.

In this problem, we have a 95% confidence level, hence$\alpha = 0.95$, z is the value of Z that has a p-value of $\frac{1+0.95}{2} = 0.975$, so the critical value is z = 1.96.

The estimate and the standard error are given, respectively, by:

$\overline{x} = 0, s = 0.69$

Hence, the interval has the bounds given as follows:

$\overline{x} - zs = 0 - 0.69(1.96) = -1.38$

$\overline{x} + zs = 0 + 0.69(1.96) = 1.38$

The 95% confidence interval is of -1.38 to 1.38, and 1.74 is outside the interval.

More can be learned about the z-distribution at brainly.com/question/24372153

#SPJ1