# Example For Z Test

### A herd of 1,500 steer was fed a special highprotein grain for a month. A random sample of 29 were weighed and had gained an average of 6.7 pounds. Contents

In national use, a vocabulary test is known to have a mean score of 68 and a standard deviation of 13. Because you have specified a significance level, you can look up the critical z value in Table of Appendix B before computing the statistic.

The null hypothesis of no difference will be rejected if the computed z statistic falls outside the range of –1.96 to 1.96. Where a and b are the limits of the confidence interval, is the sample mean, is the upper (or positive) z value from the standard normal table corresponding to half of the desired alpha level (because all confidence intervals are totaled), is the population standard deviation, and n is the size of the sample.

We have 99 percent confidence that the population mean of pin diameters lies between 1.12 and 1.18 inches. Note that this is not the same as saying that 99 percent of the machine pins have diameters between 1.12 and 1.18 inches, which would be an incorrect conclusion from this test.

Because surveys cost money to administer, researchers often want to calculate how many subjects will be needed to determine a population mean using a fixed confidence interval and significance level. Where n is the number of subjects needed, is the critical z value corresponding to the desired significance level, is the population standard deviation, and w is the desired confidence interval width.

How many subjects will be needed to find the average age of students at Fisher College plus or minus a year, with a 95 percent significance level and a population standard deviation of 3.5? Note that the confidence interval width is always double the “plus or minus” figure. (Source: www.slideshare.net)

Below are three examples to illustrate the concept of finding the probability of a sample mean using the normal distribution given you know the population standard deviation.1. The percent of area under the normal curve from negative infinity to 2.783 = 99.7%, meaning their class score is higher than 99.7% of all other samples of 31.

A researcher wanted to study the effects of mentoring on intelligence scores. He wanted to know as a baseline what the average intelligence of his students were relative to the general population.

At what percentile of intelligences is the average IQ of his students relative to his population ? The general framework of Hypothesis testing will be covered in future lessons.3.

A random sample of 36 customers attempted to rent a car on the website. The percent of area under the normal curve from negative infinity to 3 =99.865%.

Test determines if there is a significant difference between sample and population means. Test normally used for dealing with problems relating to large samples. (Source: www.educba.com)

Test formula computed by the Sample means minus population means divided by population standard deviation and sample size. When the sample size is more than 30 units than in that case the test must be performed.

In that case, he can use a test statistics method to obtain the results by taking a sample size say 500 from the city out of which suppose 280 are tea drinkers. Principal at school claims that students in his school are above average intelligence and a random sample of 30 students IQ scores have a mean score of 112.5 and mean population IQ is 100 with a standard deviation of 15.

The company randomly select 40 sample claim and calculate sample mean of Rs 195000 assuming a standard deviation of Claim is Rs 50000 and set alpha as 0.05. As per test results, we can see that 1.897 is greater than the rejection region of 1.65 so the company fails to accept the null hypothesis and the insurance company should be concerned about their current policies.

Once the above steps are performed z test statistics results are calculated. Test is used to compare the average of a normal random variable to a specified value.

Test is best on the assumption that the distribution of sample mean is normal. Test is one of the bases of statistical hypothesis testing methods and often learn at an introductory level. (Source: www.slideshare.net)

Some time z tests can be used where the data is generated from other distribution, such as binomial and Poisson. Here we discuss How to Calculate Test Statistics along with practical examples.

Array: The given set of values for which the hypothesized sample mean is to be tested. X: The hypothesized sample mean which is required to test.

Sigma: This is an optional argument which represents the population standard deviation. This TEST function in Excel gives the one-tailed probability value of a test.

The formula is given below for calculating the two-tailed P-value of a TEST for given hypothesized population means which is 5. While using the Test, we test a null hypothesis which states that the mean of the two population is equal.

Where H is called an alternative hypothesis, the mean of two populations is not equal. Let’s take an example to understand the usage of two sample Test. (Source: www.slideshare.net)

In Output Range, Select cell where you want to see the result. Thus, means of both the populations don’t differ significantly.

Test is only applicable for two samples when the variance of both the population is known. Error: If the Sigma argument value is equal to zero.

#N/A error: If the dataset values or passed array is empty. The sigma is not given and the standard deviation is zero of the passed array.

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The error in the mean is calculated by dividing the dispersion by the square root of the number of data points. So if you want to improve the reliability of the mean value by, say, a factor of 10, you would have to get 100 times more samples. 