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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?
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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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