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Tamil Nadu Board of Secondary EducationHSC Commerce Class 12

The mean breaking strength of cables supplied by a manufacturer is 1,800 with a standard deviation of 100. By a new technique in the manufacturing process it is claimed that the break - Business Mathematics and Statistics

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Question

The mean breaking strength of cables supplied by a manufacturer is 1,800 with a standard deviation of 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?

Sum

Solution

Sample size n = 50

Sample mean `bar(x)` = 1800

Sample SD S = 100

Population mean µ = 1850

Null hypothesis H0: µ = 1850

Level of significance µ = 0.01

Test statistic:

z = `(bar(x) - mu)/(sigma/sqrt("n")) "N" ∼ (0.1)`

z = `(1800 - 1840)/((100/sqrt(50))` 

= `(-50)/sqrt(100/(7.071)`

= `(-50 xx 7.0711)/100`

= `7.0711/2`

= – 3.5355

∴ z = – 3.536

Calculated value |z| = 3.536

Critical value at 1% level of significance is `"Z"_(("a")/2)` = 2.58

Inference:

Since the calculated value is greater than table

i.e. `"Z"_(("a")/2)` at 1% level of significance, the null hypothesis is rejected

Therefore we conclude that we is rejected.

Therefore we conclude that we can not support we conclude that we can support the claim of 0.01 of significance.

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Hypothesis Testing
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Chapter 8: Sampling techniques and Statistical Inference - Exercise 8.2 [Page 195]

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Samacheer Kalvi Business Mathematics and Statistics [English] Class 12 TN Board
Chapter 8 Sampling techniques and Statistical Inference
Exercise 8.2 | Q 17 | Page 195
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