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5.1 · Q4

Q.A shoe maker company produces a specific model of shoes having 15 months average lifetime. One of the employees in their R & D division claims to have developed a product that lasts longer. This latest product was worn by 30 people and lasted on average for 17 months. The variability of the original shoe is estimated based on the standard deviation of the new group which is 5.5 months. Is the designer's claim of a better shoe supported by the findings of the trial? Make your decision using two tailed testing using a level of significance of p<.05p < .05.

Andaman Nicobar CbseNCERTSubjective· 5mImportance★★★★★
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A two-tailed one-sample tt-test gives t=1.99t=1.99; as ∣t∣<2.045|t|<2.045 we cannot reject H0H_0, so the designer's claim is not statistically supported.

t=xˉ−μs/nt = \frac{\bar{x}-\mu}{s/\sqrt{n}}

  • xˉ=17\bar{x}=17 (new-shoe mean life), μ=15\mu=15 (old average), s=5.5s=5.5, n=30n=30.
  1. Hypotheses (decision made by two-tailed testing as instructed):

H0:μ=15H1:μ≠15H_0:\mu=15 \qquad H_1:\mu\neq 15

  1. Standard error: sn=5.530=5.55.477=1.0042\dfrac{s}{\sqrt{n}}=\dfrac{5.5}{\sqrt{30}}=\dfrac{5.5}{5.477}=1.0042.
  2. Test statistic:

t=17−151.0042=21.0042=1.992.t=\frac{17-15}{1.0042}=\frac{2}{1.0042}=1.992.

  1. Critical value: d.f. =n−1=29=n-1=29; two-tailed α=0.05\alpha=0.05 (0.0250.025 per tail) ⇒tcrit=±2.045\Rightarrow t_{crit}=\pm 2.045. …

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