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Q.Ten cartons are taken at random from an automatic packing machine. The mean net weight of the ten cartons is 11⋅811 \cdot 8 kg and standard deviation is 0⋅150 \cdot 15 kg. Does the sample mean differ significantly from the intended mean of 12 kg ? [Given that for d.f. = 9, t0⋅05=2⋅26t_{0 \cdot 05} = 2 \cdot 26]

CBSECBSE Class XII Board 2022Subjective· 3mImportance★★★★★
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H0:μ=12H_0:\mu=12. The calculated ∣t∣=4.0|t|=4.0 exceeds the critical value 2.262.26 at 99 d.f., so we reject H0H_0: the sample mean differs significantly from 12 kg.

One-sample tt-test:   t=xˉ−μs/n−1\;t=\dfrac{\bar{x}-\mu}{s/\sqrt{n-1}}, with degrees of freedom =n−1=n-1, where xˉ=\bar{x}= sample mean, μ=\mu= hypothesised (intended) mean, s=s= sample standard deviation, and n=n= sample size.

  1. State the hypotheses: H0:μ=12H_0:\mu=12 kg (no real difference) versus H1:μ≠12H_1:\mu\neq 12 kg (two-tailed).
  2. Given: xˉ=11.8\bar{x}=11.8, μ=12\mu=12, s=0.15s=0.15, n=10n=10, so d.f. =n−1=9=n-1=9.
  3. Standard error: sn−1=0.159=0.153=0.05\dfrac{s}{\sqrt{n-1}}=\dfrac{0.15}{\sqrt{9}}=\dfrac{0.15}{3}=0.05.
  4. Test statistic: t=11.8−120.05=−0.20.05=−4.0t=\dfrac{11.8-12}{0.05}=\dfrac{-0.2}{0.05}=-4.0, so ∣t∣=4.0|t|=4.0.
  5. Compare with the critical value: ∣t∣=4.0>t0.05=2.26|t|=4.0 > t_{0.05}=2.26. …

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