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5.1 · Q5

Q.An electric light bulbs manufacturer claims that the average life of their bulb is 2000 hours. A random sample of bulbs is tested and the life

(x) in hours recorded. The following were the outcomes: Σx=127808\Sigma x = 127808 and Σ(xˉ−x)2=9694.6\Sigma(\bar{x} - x)^2 = 9694.6 Is there sufficient evidence, at the 1% level, that the manufacturer is over estimating the life span of light bulbs?
Puducherry CbseNCERTSubjective· 5mImportance★★★★★
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With n=64n=64, xˉ=1997\bar{x}=1997 and s=12.41s=12.41, a left-tailed tt-test gives t=−1.93t=-1.93; as ∣t∣<2.39|t|<2.39 we cannot reject H0H_0 at the 1% level.

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

  • μ=2000\mu=2000 (claimed mean life), ∑x=127808\sum x=127808, ∑(xˉ−x)2=9694.6\sum(\bar{x}-x)^2=9694.6.
  1. Find nn: the average life is near 2000, and 1278082000≈63.9\dfrac{127808}{2000}\approx 63.9, so n=64n=64 (then xˉ\bar{x} is a whole number). Check: xˉ=12780864=1997\bar{x}=\dfrac{127808}{64}=1997 hours. ✓\checkmark
  2. Hypotheses ("over-estimating" ⇒\Rightarrow true mean is lower ⇒\Rightarrow left-tailed):

H0:μ=2000H1:μ<2000H_0:\mu=2000 \qquad H_1:\mu<2000

  1. Sample standard deviation:

s=9694.664−1=9694.663=153.88=12.405.s=\sqrt{\frac{9694.6}{64-1}}=\sqrt{\frac{9694.6}{63}}=\sqrt{153.88}=12.405.

  1. Standard error: sn=12.40564=12.4058=1.5506\dfrac{s}{\sqrt{n}}=\dfrac{12.405}{\sqrt{64}}=\dfrac{12.405}{8}=1.5506.
  2. Test statistic: …

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