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Q.Ten students are selected at random from a college and their heights (in cm) are found to be 100, 104, 108, 110, 118, 120, 122, 124, 126 and 128. In the light of the data, discuss the conclusion that the mean height of the students of the college is 110 cm. [Given : t9(0.05)=2.262t_9 (0.05) = 2.262]

CBSECBSE Class XII Board 2024Subjective· 3mImportance★★★★★
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With xˉ=116\bar{x}=116, ∑(x−xˉ)2=864\sum(x-\bar{x})^2=864 and n=10n=10, the test statistic ∣t∣≈1.94<2.262|t|\approx 1.94<2.262, so H0H_0 (mean =110=110 cm) is accepted.

One-sample tt-test: t=xˉ−μs/nt=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}, where s2=1n−1∑(xi−xˉ)2s^2=\dfrac{1}{n-1}\sum (x_i-\bar{x})^2, with n−1n-1 degrees of freedom. Accept H0H_0 if ∣t∣<tn−1(0.05)|t|<t_{n-1}(0.05).

  1. Hypotheses: H0:μ=110H_0:\mu=110 cm (no significant difference); H1:μ≠110H_1:\mu\neq 110 cm.
  2. Sample mean: xˉ=100+104+108+110+118+120+122+124+126+12810=116010=116.\bar{x}=\dfrac{100+104+108+110+118+120+122+124+126+128}{10}=\dfrac{1160}{10}=116.
  3. Deviations di=xi−116d_i=x_i-116: −16,−12,−8,−6,2,4,6,8,10,12-16,-12,-8,-6,2,4,6,8,10,12. Squares: 256,144,64,36,4,16,36,64,100,144256,144,64,36,4,16,36,64,100,144, summing to ∑di2=864.\sum d_i^2=864.
  4. Sample standard deviation: s2=864n−1=8649=96⇒s=96≈9.798.s^2=\dfrac{864}{n-1}=\dfrac{864}{9}=96\Rightarrow s=\sqrt{96}\approx 9.798. …

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