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The Z-test is a statistical procedure used to determine if there is a significant difference between the means of a sample and a population. In our exercise, the question explores whether the average cost of building a home differs significantly from an assumed population mean of \( \\( 117.91 \) per square foot. The Z-test is particularly useful when the sample size is large, typically over 30, allowing the Central Limit Theorem to apply.

Here's how we apply the Z-test:

• First, ensure the sample size is sufficiently large, enabling the use of the normal distribution approximation.
• Identify your sample mean \( (\bar{x}) \), population mean \( (\mu) \), sample standard deviation \( (\sigma) \), and sample size \( (n) \).
• Use the formula \[ Z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \] to calculate the Z value.
• Use the standard normal distribution to evaluate if the calculated Z value indicates a statistically significant difference.

By calculating the Z value for our sample, we determine if the observed mean significantly deviates from the expected mean of \( \\) 117.91 \). This helps us make informed decisions based on statistical evidence rather than assumptions.

In hypothesis testing, formulating clear hypotheses is crucial. The null hypothesis \( (H_0) \) and alternative hypothesis \( (H_a) \) serve as starting points for the test.

For the given exercise:

• The null hypothesis \( (H_0) \) is formulated as \( \mu = 117.91 \). This suggests that there is no significant difference; the mean cost of building a home is \( \$ 117.91 \).
• The alternative hypothesis \( (H_a) \) posits \( \mu eq 117.91 \). It suggests that the mean cost differs, opening up the possibility for either a higher or lower average than what is reported.

In this context, we conduct a two-tailed test since we are interested in any significant deviation away from \( 117.91 \), irrespective of direction. The decision to reject or not the null hypothesis is based on the test results and helps determine if there is substantial evidence for any noted disparity in the mean costs.

The significance level \( (\alpha) \) in hypothesis testing defines the probability of rejecting the null hypothesis when it is actually true. This level, often expressed as a percentage, reflects the risk analysts are willing to take on reaching a false positive conclusion.

In the context of the exercise:

• We use a significance level of \( 0.10 \). In practical terms, this means that there is a 10% risk of incorrectly rejecting the null hypothesis.
• For a two-tailed test at this significance level, the critical Z-values designate the cutoff points beyond which we reject \( H_0 \). In this case, values are approximately \( \pm 1.645 \).

This level guides our understanding of the confidence we have in the test results. For our exercise, since the calculated Z fell within the critical range, we did not reject the null hypothesis, indicating we lack enough evidence at the 0.10 significance level to state that the mean differs from \( \$ 117.91 \). This decision helps ensure that any conclusions drawn are robust and statistically justified.