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 sample data is \(x_1 = 115.7\), \(x_2=62.9\), \(x_3 = 168.5\), \(x_4=135.7\).
The sample mean \(\bar{x}=\frac{115.7 + 62.9+168.5 + 135.7}{4}=\frac{482.8}{4}=120.7\)

Step2: Calculate the z - statistic

The formula for the z - statistic in a one - sample z - test (when population standard deviation \(\sigma\) is known) is \(z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\)
We are given \(\mu = 86.7\), \(\sigma=40.4\), \(n = 4\)
Substitute the values into the formula:

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Step3: Determine the critical value and make a decision

For a two - tailed test with a significance level of \(\alpha = 0.05\), the critical values are \(z_{\alpha/2}=\pm1.96\)
Since \(|z|=1.68<1.96\), we fail to reject the null hypothesis.

Answer:

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