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?
Advertisements
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 critical value
Define the level of significance
What is single tailed test
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:
Errors in sampling are of
Choose the correct alternative:
Type I error is
Choose the correct alternative:
Type II error is
Explain in detail the test of significance of a single mean
