Sagot :
Using the z-distribution, it is found that:
- The 95% confidence interval is of -1.38 to 1.38.
- The value of the sample mean difference is of 1.74, which falls outside the 95% confidence interval.
What is the z-distribution confidence interval?
The confidence interval is:
[tex]\overline{x} \pm zs[/tex]
In which:
- [tex]\overline{x}[/tex] is the difference between the population means.
- s is the standard error.
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 by:
[tex]\overline{x} = 0, s = 0.69[/tex]
Hence the bounds of the interval are given by:
[tex]\overline{x} - zs = 0 - 1.96(0.69) = -1.38[/tex]
[tex]\overline{x} + zs = 0 + 1.96(0.69) = 1.38[/tex]
1.74 is outside the interval, hence:
- The 95% confidence interval is of -1.38 to 1.38.
- The value of the sample mean difference is of 1.74, which falls outside the 95% confidence interval.
More can be learned about the z-distribution at https://brainly.com/question/25890103
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