n= 25 scores from a population with mean =20
Chat with our AI personalities
Depending on whether you subtract actual value from expected value or other way around, a positive or negative percent error, will tell you on which side of the expected value that your actual value is. For example, suppose your expected value is 24, and your actual value is 24.3 then if you do the following calculation to figure percent error:[percent error] = (actual value - expected value)/(actual value) - 1 --> then convert to percent.So you have (24.3 - 24)/24 -1 = .0125 --> 1.25%, which tells me the actual is higher than the expected. If instead, you subtracted the actual from the expected, then you would get a negative 1.25%, but your actual is still greater than the expected. My preference is to subtract the expected from the actual. That way a positive error tells you the actual is greater than expected, and a negative percent error tells you that the actual is less than the expected.
Well, sort of. The Chi-square distribution is the sampling distribution of the variance. It is derived based on a random sample. A perfect random sample is where any value in the sample has any relationship to any other value. I would say that if the Chi-square distribution is used, then every effort should be made to make the sample as random as possible. I would also say that if the Chi-square distribution is used and the sample is clearly not a random sample, then improper conclusions may be reached.
The simple answer is no. This depends on a lot of factors such as alpha which determines the critical value and the absolute value of the difference between the claim and sample data. Mathematically speaking, all things being equal, the larger the sample size the larger the absolute value of the test statistic. The formula for the test statistic mean with sigma known is shown below. You can substitute values in and perform the mathematics. The larger the sample size, the larger the Z value; but note if the numerator is small, even a small denominator will not produce a large Z value. In fact, the numerator could be zero which would make the test statistic zero. Z = (Xbar - μxbar)/(σ/√n) (formula from Elementary Statistics by Mario F. Triola)
to produce a product with zero defects
the larger you r sample size the better your estimate. imagine take the height of person to estimate the average high of an adult male. would one person's height be a good estimate, or would taking the average height of 100, or 5000 adult males will produce a better estimate?