a local weather enthusiast wants to see if the last five summers have been warmer than normal for eloy…

a local weather enthusiast wants to see if the last five summers have been warmer than normal for eloy, arizona. from the national oceanic and atmospheric administration (noaa), he knows the 30 - year average for july was 90° with a standard deviation of 2.1°. the last five summers recorded have an average of 93.5°. if he uses a = 5%, which of the following would be valid based on the data he has gathered?\nhe can conclude that the summers have been warmer than the 30 - year average.\nhe can conclude that the summers will continue to be warmer for the next 10 years.\nhe does not have enough information to make a decision.\nthe average for the last five years fits with the 30 - year average.

a local weather enthusiast wants to see if the last five summers have been warmer than normal for eloy, arizona. from the national oceanic and atmospheric administration (noaa), he knows the 30 - year average for july was 90° with a standard deviation of 2.1°. the last five summers recorded have an average of 93.5°. if he uses a = 5%, which of the following would be valid based on the data he has gathered?\nhe can conclude that the summers have been warmer than the 30 - year average.\nhe can conclude that the summers will continue to be warmer for the next 10 years.\nhe does not have enough information to make a decision.\nthe average for the last five years fits with the 30 - year average.

Answer

Explanation:

Step1: Calculate the z - score

We use the z - score formula $z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}$, where $\bar{x} = 93.5$ (sample mean), $\mu = 90$ (population mean), $\sigma=2.1$ (population standard deviation), and $n = 5$ (sample size). $z=\frac{93.5 - 90}{\frac{2.1}{\sqrt{5}}}=\frac{3.5}{\frac{2.1}{\sqrt{5}}}\approx\frac{3.5}{0.94}\approx3.72$

Step2: Determine the critical value

For a one - tailed test with $\alpha = 0.05$, the critical z - value from the standard normal distribution table is approximately $z_{0.05}=1.645$.

Step3: Make a decision

Since the calculated z - score ($z\approx3.72$) is greater than the critical z - value ($z_{0.05}=1.645$), we reject the null hypothesis (that the sample comes from a population with mean equal to the 30 - year average).

Answer:

He can conclude that the summers have been warmer than the 30 - year average.