Dear ---Bilal,

The following text gives an example of the requested calculations: for more details please use the following link:

https://en.wikipedia.org/wiki/Z-test

Example

Suppose that in a particular geographic region, the mean and standard deviation of scores on a reading test are 100 points, and 12 points, respectively. Our interest is in the scores of 55 students in a particular school who received a mean score of 96. We can ask whether this mean score is significantly lower than the regional mean—that is, are the students in this school comparable to a simple random sample of 55 students from the region as a whole, or are their scores surprisingly low?

We begin by calculating the standard error of the mean:

SE = sigma/root n = 12/root 55= 12/7.42 =1.62

where is the population standard deviation

Next we calculate the z-score, which is the distance from the sample mean to the population mean in units of the standard error:

z = M-u/SE = 96-100//1.62= -2.47

In this example, we treat the population mean and variance as known, which would be appropriate if all students in the region were tested. When population parameters are unknown, a t test should be conducted instead.

The classroom mean score is 96, which is −2.47 standard error units from the population mean of 100. Looking up the z-score in a table of the standard normal distribution, we find that the probability of observing a standard normal value below −2.47 is approximately 0.5 − 0.4932 = 0.0068. This is the one-sided p-value for the null hypothesis that the 55 students are comparable to a simple random sample from the population of all test-takers. The two-sided p-value is approximately 0.014 (twice the one-sided p-value).

Another way of stating things is that with probability 1 − 0.014 = 0.986, a simple random sample of 55 students would have a mean test score within 4 units of the population mean. We could also say that with 98.6% confidence we reject the null hypothesis that the 55 test takers are comparable to a simple random sample from the population of test-takers.

The Z-test tells us that the 55 students of interest have an unusually low mean test score compared to most simple random samples of similar size from the population of test-takers. A deficiency of this analysis is that it does not consider whether the effect size of 4 points is meaningful. If instead of a classroom, we considered a subregion containing 900 students whose mean score was 99, nearly the same z-score and p-value would be observed. This shows that if the sample size is large enough, very small differences from the null value can be highly statistically significant. See statistical hypothesis testing for further discussion of this issue.

Hoping this will be helpful,

Rafik

Dear ---Bilal,

The following text gives an example of the requested calculations: for more details please use the following link:

https://en.wikipedia.org/wiki/Z-test

Example

Suppose that in a particular geographic region, the mean and standard deviation of scores on a reading test are 100 points, and 12 points, respectively. Our interest is in the scores of 55 students in a particular school who received a mean score of 96. We can ask whether this mean score is significantly lower than the regional mean—that is, are the students in this school comparable to a simple random sample of 55 students from the region as a whole, or are their scores surprisingly low?

We begin by calculating the standard error of the mean:

SE = sigma/root n = 12/root 55= 12/7.42 =1.62

where is the population standard deviation

Next we calculate the z-score, which is the distance from the sample mean to the population mean in units of the standard error:

z = M-u/SE = 96-100//1.62= -2.47

In this example, we treat the population mean and variance as known, which would be appropriate if all students in the region were tested. When population parameters are unknown, a t test should be conducted instead.

The classroom mean score is 96, which is −2.47 standard error units from the population mean of 100. Looking up the z-score in a table of the standard normal distribution, we find that the probability of observing a standard normal value below −2.47 is approximately 0.5 − 0.4932 = 0.0068. This is the one-sided p-value for the null hypothesis that the 55 students are comparable to a simple random sample from the population of all test-takers. The two-sided p-value is approximately 0.014 (twice the one-sided p-value).

Another way of stating things is that with probability 1 − 0.014 = 0.986, a simple random sample of 55 students would have a mean test score within 4 units of the population mean. We could also say that with 98.6% confidence we reject the null hypothesis that the 55 test takers are comparable to a simple random sample from the population of test-takers.

The Z-test tells us that the 55 students of interest have an unusually low mean test score compared to most simple random samples of similar size from the population of test-takers. A deficiency of this analysis is that it does not consider whether the effect size of 4 points is meaningful. If instead of a classroom, we considered a subregion containing 900 students whose mean score was 99, nearly the same z-score and p-value would be observed. This shows that if the sample size is large enough, very small differences from the null value can be highly statistically significant. See statistical hypothesis testing for further discussion of this issue.

Hoping this will be helpful,

Rafik

Thanks a lot sir

pVALUE

Suppose that p = 0.05, how to obtain Z?

Recall that p = 1 - F(Z)

With known pValue p = 0.05, then:

0.05 = 1 - F(Z)

0.05 + F(Z) = 1

F(Z) = 1 - 0.05

F(Z) = 0.95

Now go to the Z Table, the table is listed by 2 columns: Z column and (1 - a) column. Read along the (1 - a) column down to the value 0.95, the corresponding value Z column is 1.65. This is the Z critical value for 0.95 at 0.05 pValue.

Z SCORE is determine by the followings:

Z = (Xobs - X^) / S

Xobs = the value to be evaluates

X^ = mean of the observed series

S = standard deviation of the observed series

Suppose with have the following observed values (1,2,3,4,5), and we want to determine the Z score for 4 or Xobs = 4, what is the Z score?

X^ = (1 + 2 + 3 + 4 = 5) / 5

X^ = 3.00

S = 1.58

Now calculate the Z score for Xobs = 4

Z = (Xobs - X^) / S

Z = (4 - 3) / 1.58

Z = 1 / 1.58

Z = 0.63

NOW, go to the Z table to find the % probability for Z = 0.63 or at Z = 0.63, what is F(Z) = 1 - a ?

Looking at the Z table, at Z = 0.63, the corresponding percentage probability is 0.736 or 73.6%.

What is the pValue in this example?

pValue = 1 - F(Z)

where F(Z) = 0.736, then 1 - 0.736 = 0.264. Thus, pValue = 0.264 where if the expected value is pValue = 0.05, then pValue(obs) = 0.264 fails or it is NOT significant because 0.264 > 0.05. Significant is found only where pValue(obs) < 0.05.