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Worked Examples · Example 5
Q.

Let us consider the average rainfall in a given area is 8 inches. However, a local meteorologist 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:

Year20162017201820192020
Rainfall (inches)85756
Puducherry CbseNCERTSubjective· 5mImportance★★★★★
23% · 3/13 Questions
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A one-sample two-tailed tt-test on the five yearly rainfalls gives t=−3.09t=-3.09; as ∣t∣>2.776|t|>2.776 we reject H0H_0, so the period's rainfall differs significantly from 8 inches.

t=xˉ−μs/n,s=∑(x−xˉ)2n−1t = \frac{\bar{x}-\mu}{s/\sqrt{n}}, \qquad s=\sqrt{\frac{\sum (x-\bar{x})^2}{n-1}}

  • xˉ\bar{x} = sample mean, μ\mu = claimed population mean =8=8
  • ss = sample standard deviation, nn = sample size =5=5
  1. Hypotheses (μ=8\mu=8 is the overall average; "significantly different" ⇒\Rightarrow two-tailed):

H0:μ=8H1:μ≠8H_0:\mu=8 \qquad H_1:\mu\neq 8

  1. Working table (data 8,5,7,5,68,5,7,5,6):
Yearxxx−xˉx-\bar{x}(x−xˉ)2(x-\bar{x})^2
201681.83.24
20175-1.21.44
201870.80.64
20195-1.21.44
20206-0.20.04
Sum3106.80
  1. Sample mean: xˉ=315=6.2\bar{x}=\dfrac{31}{5}=6.2 inches.
  2. Sample standard deviation: s=6.805−1=1.7=1.304s=\sqrt{\dfrac{6.80}{5-1}}=\sqrt{1.7}=1.304.
  3. Test statistic: …

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