Question

in juneau, alaska, the 30-year annual snowfall average is 86.7 inches with a standard deviation of 40.4 inches. the last four years saw an average annual snowfall of 115.7 inches, 62.9 inches, 168.5 inches, and 135.7 inches. hia performs a hypothesis test on this data to determine if the next 30-year norm will have a different average if the trend from the last four years continues. she uses a significance level of 5%. which of the following is a conclusion that she may make?

the z-statistic is 1.44, so the null hypothesis cannot be rejected.

the z-statistic is 1.68, so the null hypothesis cannot be rejected.

the z-statistic is 1.85, so the null hypothesis should be rejected.

the z-statistic is 4.6, so the null hypothesis should be rejected.

Explanation:

Step1: Calculate the sample mean

The last four years' snowfall: 115.7, 62.9, 168.5, 135.7.
Sample mean $\bar{x}=\frac{115.7 + 62.9 + 168.5 + 135.7}{4}=\frac{482.8}{4} = 120.7$ inches.

Step2: Identify population parameters

Population mean $\mu = 86.7$ inches, standard deviation $\sigma = 40.4$ inches, sample size $n = 4$.

Step3: Calculate the z - statistic

The formula for the z - statistic in a hypothesis test for the mean (when $\sigma$ is known) is $z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}$.
Substitute the values:
$\frac{\sigma}{\sqrt{n}}=\frac{40.4}{\sqrt{4}}=\frac{40.4}{2}=20.2$
$z=\frac{120.7 - 86.7}{20.2}=\frac{34}{20.2}\approx1.68$

Step4: Determine critical value

For a two - tailed test with $\alpha = 0.05$, the critical $z$ - values are $\pm1.96$.

Step5: Compare z - statistic and critical value

Since $|1.68|\lt1.96$, we fail to reject the null hypothesis.

Answer:

The z - statistic is 1.68, so the null hypothesis cannot be rejected.