Posted by **Anonymous** on Saturday, August 11, 2012 at 11:38pm.

For normal population mean 60, standard deviation 15, after treatment is given to the member of sample , the sample mean is 65, and sample size is 25.

is the sample mean sufficient to conclude that the treatment has significant effect? Give reason for decision?

- statistics -
**PsyDAG**, Sunday, August 12, 2012 at 6:51pm
Z = (mean1 - mean2)/standard error (SE) of difference between means

SEdiff = √(SEmean1^2 + SEmean2^2)

SEm = SD/√n

If only one SD is provided, you can use just that to determine SEdiff.

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to the Z score. Is that less than the level of significance you are using?

## Answer This Question

## Related Questions

- Statistics - . A random sample is selected from a normal population with a mean ...
- Probabilty & Statistic - A random sample is selected from a normal population ...
- statistics - An IQ is designed so that the mean is 100 and the standard ...
- Statistics - An IQ test is designed so that the mean is 100 and the standard ...
- statistics - The scores of students on the ACT college entrance examination in ...
- statistics - what are the mean and standard deviation of a sampling distribution...
- Math (Statistic) - Considered the sampling distribution of a sample mean ...
- Statistics - he Oil Price Information Center reports the mean price per gallon ...
- statistics - A random sample of size 20 from a normal population will sample ...
- Statistics - A random sample of size 20 from a normal population will sample ...

More Related Questions