Respuesta :
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:
[tex]\overline{x} \pm zs[/tex]
In which:
- [tex]\overline{x}[/tex] 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[tex]\alpha = 0.95[/tex], z is the value of Z that has a p-value of [tex]\frac{1+0.95}{2} = 0.975[/tex], so the critical value is z = 1.96.
The estimate and the standard error are given, respectively, by:
[tex]\overline{x} = 0, s = 0.69[/tex]
Hence, the interval has the bounds given as follows:
[tex]\overline{x} - zs = 0 - 0.69(1.96) = -1.38[/tex]
[tex]\overline{x} + zs = 0 + 0.69(1.96) = 1.38[/tex]
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 https://brainly.com/question/24372153
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