Solutions

Problem Part Solution
1 - - Symmetrical
- Mean of 0
- Exact shape depends on the degrees of freedom
- Bell Shaped
- More area in the tails than the standard normal distribution
2 - \(df = n-1 \text{(that is sample size} - 1\))
3 - \[\bar{x}-t^* \times \frac{s}{\sqrt{n}}, \bar{x} + t^* \times \frac{s}{\sqrt{n}}\]
4 - - Sigma is not known
- We compute a t-score from the Student t-distribution rather than a z-score from the normal distribution. (We cannot use the Normal Applet to compute t-scores.)
- We must now consider degrees of freedom for the t-distribution where the normal distribution did not have degrees of freedom.
5 - \[t=\frac{\bar{x}-\mu}{s/\sqrt{n}}\]
6 - - A simple random sample was drawn from a population.
- \(\bar{x}\) is normally distributed.
7 - The mean is 46.733. The standard devation is 8.827
8 - It is a random sample from a population. The \(\bar{x}\)’s are normally distributed
9 - (41.845 , 51.622)
10 - (39.948 , 53.518)
11 A (41.845 , 51.622)
11 B (39.948 , 53.518)
12 - The margin of error for the confidence interval is smaller for a 95% confidence level than a 99% confidence level.
13 A \(H_o: \mu = 47\)
\(H_a: \mu \neq 47\)
13 B t = -0.117
13 C df = 14
13 D Solution
13 E P-value = 0.909
13 F fail to reject the null hypothesis
13 G We have insufficient evidence to conclude that the mean age of this realtor’s customers who are buying second homes, is different than the national average.