A simple random sample of 16 birth weights to mothers given a special vitamin supplement during pregnancy has standard deviation of 0.959 kg. Test the claim that this sample comes for a population with a standard deviation equal to 0.470 kg. Use a 0.05 significance level. State the initial and final conclusion. Could anyone show me how to solve this?

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1 Answer

Standard error=0.959/√16=0.23975kg.

Critical value fot 0.05 significance level with 15 degrees of freedom is 2.131. Apply this to the standard error to get the margin of error: 2.131×0.23975=0.5109 approx.

Therefore the range for the standard deviation is 0.959±0.511: 0.448 to 1.470kg. The value 0.470kg is within this range so the claim is valid (95% certain).

Initial conclusion might have been that the sample standard deviation differed from the population standard deviation; but the final conclusion is that the population standard deviation is within the margin of error as applied to the sample, so the sample analysis shows that the standard deviation is not significantly different from that of the population.

by Top Rated User (711k points)

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