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Q.Consider the following hypothesis : H0:μ=35H_0 : \mu = 35 H1:μ≠35H_1 : \mu \neq 35 A sample of 81 items is taken whose mean is 37⋅537 \cdot 5 and the standard deviation is 5. Test the hypothesis at 5% level of significance. [Given : Critical value of Z for a two-tailed test at 5% level of significance is 1⋅961 \cdot 96]

CBSECBSE Class XII Board 2022Subjective· 2mImportance★★★★★
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The test statistic Z=4.5Z=4.5 exceeds the critical value 1.961.96, so H0:μ=35H_0:\mu=35 is rejected at the 5% level of significance.

Z=xˉ−μs/nZ=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}, where xˉ\bar{x} = sample mean, μ\mu = hypothesised population mean, ss = standard deviation, nn = sample size.

  1. State the hypotheses: H0:μ=35H_0:\mu=35 versus H1:μ≠35H_1:\mu\neq 35 (a two-tailed test).
  2. Given data: n=81n=81, xˉ=37.5\bar{x}=37.5, s=5s=5, μ=35\mu=35.
  3. Standard error =sn=581=59≈0.5556=\dfrac{s}{\sqrt{n}}=\dfrac{5}{\sqrt{81}}=\dfrac{5}{9}\approx 0.5556.
  4. Test statistic Z=xˉ−μs/n=37.5−355/9=2.50.5556=2.5×95=22.55=4.5Z=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{37.5-35}{5/9}=\dfrac{2.5}{0.5556}=2.5\times\dfrac{9}{5}=\dfrac{22.5}{5}=4.5. …

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