Advertisements
Advertisements
प्रश्न
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?
उत्तर
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.
APPEARS IN
संबंधित प्रश्न
What is null hypothesis? Give an example
Define alternative hypothesis
Define the level of significance
What is a type I error?
A sample of 100 items, draw from a universe with mean value 4 and S.D 3, has a mean value 63.5. Is the difference in the mean significant at 0.05 level of significance?
Choose the correct alternative:
Type I error is
Choose the correct alternative:
Type II error is
Explain the procedures of testing of hypothesis
Explain in detail the test of significance of a single mean
The mean I.Q of a sample of 1600 children was 99. Is it likely that this was a random sample from a population with a mean I.Q 100 and standard deviation of 15? (Test at 5% level of significance)