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Question 39 of 39
Q.
  1. The mean breaking strength of cables supplied by a manufacturer is 1,800 with a Standard Deviation 100. By a new technique in the manufacturing process, it is claimed that the breaking strength of the cables has increased. In order to test this claim a sample of 50 cables is tested. It is found that the mean breaking strength is 1,850. Can you support the claim at 0.01 level of significance ? OR
  2. Calculate the seasonal indices from the following data using Simple Average method.
YearI QuarterII QuarterIII QuarterIV Quarter
200872686276
200978747872
201074707276
201176747472
201272727668
Puducherry TnboardTamil Nadu HSC (DGE) Commerce Board 2026Subjective· 5mImportance★★★★★
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(a) Right-tailed ZZ-test: Z=3.54Z=3.54 exceeds the 1%1\% critical value 2.332.33 → reject H0H_0, claim supported. (b) Quarter averages 74.4,71.6,72.4,72.874.4,71.6,72.4,72.8; grand average 72.872.8; indices =Q avg72.8×100=\frac{\text{Q avg}}{72.8}\times100.

Part (a) — Z-test for a single mean

Given μ=1800, σ=100, n=50, xˉ=1850\mu=1800,\ \sigma=100,\ n=50,\ \bar x=1850, level α=0.01\alpha=0.01.

Step 1 — Hypotheses (claim: strength increased):

H0:μ=1800vsH1:μ>1800 (right-tailed).H_0:\mu=1800\quad\text{vs}\quad H_1:\mu>1800\ \text{(right-tailed)}.

Step 2 — Test statistic:

Z=xˉ−μσ/n=1850−1800100/50=50100/7.071=5014.142=3.54.Z=\frac{\bar x-\mu}{\sigma/\sqrt{n}}=\frac{1850-1800}{100/\sqrt{50}}=\frac{50}{100/7.071}=\frac{50}{14.142}=3.54.

Step 3 — Decision. Critical value at 1%1\% (one-tailed) =2.33=2.33. Since calculated ∣Z∣=3.54>2.33|Z|=3.54>2.33, we reject H0H_0.

The sample gives strong evidence that the mean breaking strength has increased.

Part (b) — Seasonal indices by the simple-average method

Step 1 — Quarter totals and averages (5 years):

QuarterTotalAverage
I37274.4
II35871.6
III36272.4
IV36472.8

Step 2 — Grand (overall) average of the quarterly averages:

yˉˉ=74.4+71.6+72.4+72.84=291.24=72.8.\bar{\bar{y}}=\frac{74.4+71.6+72.4+72.8}{4}=\frac{291.2}{4}=72.8.

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