Let us consider the average rainfall in a given area is 8 inches. However, a local meteorological department claims that rainfall was above average from 2016-2020 and argues that average rainfall during this period was significantly different from overall average rainfall. The following is the average rainfall for the observed period of 2016-2020.
|
Year |
2016 |
2017 |
2018 |
2019 |
2020 |
|
Rainfall (inches) |
8 |
5 |
7 |
5 |
6 |
Is the meteorological department's claim about the rainfall true? (Given $α = 0.025$)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Reject $H_0$; there is sufficient evidence to conclude rainfall was significantly different from the average.
Given sample size $n = 5$.
So, sample mean $\bar x=\frac{8+5+7+5+6}{5}=\frac{31}{5}= 6.2$
Sample standard deviation $S=\frac{\sqrt{(8 -6.2)^2 + (5-6.2)^2 + (7-6.2)^2 + (5-6.2)^2 + (6-6.2)^2}}{4}$
$=\sqrt{\frac{3.24 +1.44 + 0.64 + 1.44 + 0.04}{4}}=\sqrt{1.7}=1.3$
Given population mean $μ = 8$.
Let the hypothesis be given as
Null hypothesis $H_0: μ≥8$
Alternative hypothesis $H_a: μ <8$.
So, the test statistic $t =\frac{\bar x - μ_0}{\frac{S}{\sqrt{n}}}=\frac{6.2-8}{\frac{1.3}{\sqrt{5}}}=-\frac{(1.8)\sqrt{5}}{1.3}$
$⇒t=-3.096$
$df=5-1=4$, so $t_α=t_{0.025}$ at $df = 4$ is
$t_{0.025} = 2.776⇒-t_{0.025}=-2.776$
$∵t≤-tα$, so reject $H_0$.
Hence, the meteorological department's claim about the rainfall is false.