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प्रश्न
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?
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उत्तर
Sample size n = 100
Sample mean `bar(x)` = 63.5
Sample SD S = 3
Population mean µ = 64
Population SD σ = 3
Null Hypothesis H0: µ = 64 .....(the sample has been drawn from the population mean µ = 64 and SD σ = 3)
Alternative Hypothesis H1: µ ≠ 64 ....(two tail)
i.e The sample has not been drawn from the population mean µ = 64 and SD σ = 3
The level of significance α = 5% = 0.05
Test statistic z = `(63.5 - 64)/(3/sqrt(100)`
= `(-0.5)/((3/10))`
= `(-0.5)/0.3`
= - 1.667
`|z| = 1.667`
∴ calculated z = 1.667
critical value at 5% level of significance is `"z"_("a"/2)` = 1.96
Inference:
At 5% level of significance `"Z" < "Z"_("a"/2)`
Since the calculated value is less than the table value the null hypothesis is accepted.
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संबंधित प्रश्न
What is null hypothesis? Give an example
Define alternative hypothesis
Define critical region
Define critical value
What is single tailed test
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?
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
