Using the z-distribution, as we are working with population data, it is found that:
The 95% confidence interval is -1.38 to 1.38, and 1.74 is outside the interval.
The confidence interval is:
[tex]\overline{x} \pm zs[/tex]
In which:
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 parameters of the interval are given by:
[tex]\overline{x} = 0, s = 0.69[/tex]
Hence, the lower and the upper bound of the interval are given, respectively, as follows:
[tex]\overline{x} - zs = 0 - 1.96(0.69) = -1.38[/tex]
[tex]\overline{x} + zs = 0 + 1.96(0.69) = 1.38[/tex]
The 95% confidence interval is -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/25890103
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