in juneau, alaska, the 30 - year annual snowfall average is 86.7 inches with a standard deviation of 40.4…

Answer:

1. First, calculate the sample mean $\bar{x}$ of the last four - year snowfall:
   • $\bar{x}=\frac{115.7 + 62.9+168.5 + 135.7}{4}=\frac{482.8}{4}=120.7$ inches.
   • The population mean $\mu = 86.7$ inches and the population standard deviation $\sigma = 40.4$ inches, and the sample size $n = 4$.
   • The formula for the z - statistic in a one - sample z - test is $z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}$.
2. Then, substitute the values into the z - statistic formula:
   • Substitute $\bar{x}=120.7$, $\mu = 86.7$, $\sigma = 40.4$, and $n = 4$ into the formula.
   • First, calculate $\frac{\sigma}{\sqrt{n}}=\frac{40.4}{\sqrt{4}}=\frac{40.4}{2}=20.2$.
   • Then, $z=\frac{120.7 - 86.7}{20.2}=\frac{34}{20.2}\approx1.68$.
3. Next, determine the critical value for a two - tailed test with a significance level of $\alpha = 0.05$:
   • For a two - tailed test with $\alpha=0.05$, the critical values are $z_{\alpha/2}=\pm1.96$.
   • Since $|z| = 1.68<1.96$, we cannot reject the null hypothesis.

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

Explanation:

Step1: Calculate sample mean

$\bar{x}=\frac{115.7 + 62.9+168.5 + 135.7}{4}=120.7$

Step2: Identify population parameters

$\mu = 86.7$, $\sigma = 40.4$, $n = 4$

Step3: Calculate denominator of z - formula

$\frac{\sigma}{\sqrt{n}}=\frac{40.4}{\sqrt{4}} = 20.2$

Step4: Calculate z - statistic

$z=\frac{120.7 - 86.7}{20.2}\approx1.68$

Step5: Determine decision

Since $|z| = 1.68<1.96$ (critical value), cannot reject null hypothesis.