Simplify (5 + 1)2 - (11 + 32) + 4.
2 answers:
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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.
The confidence interval is:
In which:
is the difference between the population means.
- s is the standard error.
In this problem, we have a 95% confidence level, hence, z is the value of Z that has a p-value of
, so the critical value is z = 1.96.
The estimate and the standard error are given by:
Hence the bounds of the interval are given by:
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 brainly.com/question/25890103
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