n= 25 scores from a population with mean =20
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?
The chi-square goodness of fit test determines whether a set of categorical data have an expected value that is similar to the observed value. For example, if you hypothesized that each Zodiac animal sign has the same proportion of people, then you would look at a sample. You would organize that sample into a chart showing how many people fit into which Zodiac animal. This is your observed value. Then you would have another table to express what your expected value is, which is 1/12 of the total sample population (because there are 12 Zodiac animal signs). Your null hypothesis is that each Zodiac animal sign will have 1/12 of the population. Next, you verify that you can use the chi-square test. To use the chi-square test you must verify that all your expected counts are over 5. Afterwards, you need to calculate a special variable notated as x^2. For each animal, you take the ((observed-expected)^2) / expected. Then you add it up. This is your x^2. Afterwards, you find the P-Value by using a chart or a calculator. If your P-Value is small, this means that the sample could not have occurred by chance and as a result, you can reject your null hypothesis. You would have significant evidence to prove that each Zodiac animal contains a different proportion. If your p-value is large, you fail to reject your null hypothesis.
The null hypothesis in a chi-square goodness-of-fit test states that the sample of observed frequencies supports the claim about the expected frequencies. So the bigger the the calculated chi-square value is, the more likely the sample does not conform the expected frequencies, and therefore you would reject the null hypothesis. So the short answer is, REJECT!
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.
The mass will be 1,6 g.
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.
Because it is a rock, not a mineral, the Mohs value will vary significantly from sample to sample. An average range would be between 3.0 and 3.8.
Genotypes are not created by phenotypes, they are the alleles/genes of the organism. Genotypes (in combination with environment) produce phenotypes. It would be expected that the genotypes Bb and BB would produce the phenotype B.
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
No. Unless the non-financial value was more than enough to offset the expected financial loss.
I would appreciate a written confirmation from you
If you make the unjustified assumption that the 25 people were a representative sample of the 600, then you would expect 48 sick people. However, that is merely the expected value, the true number is likely to be different from 48.