Z-test Calculator for Mean Examples

Z-test Calculator for mean with examples

In this tutorial we will discuss $z$-test calculator for testing population mean with step by step numerical examples on $Z$-test for testing single population mean

Z test calculator for one mean

The $Z$-test calculator for testing population mean makes it easy to calculate the test statistic, $Z$ critical value and the $p$-value given the sample information, level of significance and the type of alternative hypothesis (i.e. left-tailed, right-tailed or two-tailed.)

Z test Calculator for mean
Population Mean ($\mu$)
Population Standard Deviation ($\sigma$)
Sample Size ($n$)
Sample Mean ($\overline{x}$)
Level of Significance ($\alpha$)
Tail : Left tailed
Right tailed
Two tailed
Results
Standard Error of Mean:
Test Statistics Z:
Z-critical value:
p-value:

How to use $z$-test calculator for testing single proportion?

Step 1 - Enter the population mean $\mu$ and population standard deviation $\sigma$.

Step 2 - Enter the sample size $n$

Step 3 - Enter the sample mean $\overline{X}$

Step 4 - Enter the level of significance $\alpha$

Step 5 - Select the alternative hypothesis (left-tailed / right-tailed / two-tailed)

Step 6 - Click on "Calculate" button to get the result

$Z$-test for testing mean Example 1

The average cost of tuition and room and board at a small private liberal arts college is reported to be \$8,500 per term, but a financial administrator believes that the average cost is lower. A study conducted using 350 small liberal arts colleges showed that the average cost per term is \$8,445. The population standard deviation is \$1,200. Let$\alpha= 0.05$. What is the test statistic for this test? Solution Let$X$denote cost of tution, room and board at a small private liberal arts college. Given that$n = 350$, sample mean cost of tution, room and board at a small private liberal arts college is$\overline{x} = 8445$, the population standard deviation is$\sigma = 1200$. Step 1 Hypothesis Testing Problem The hypothesis testing problem is$H_0 : \mu = 8500$against$H_1 : \mu < 8500$($\text{left-tailed}) Step 2 Test Statistic The test statistic for testing above hypothesis testing problem is  \begin{aligned} Z& =\frac{\overline{x} -\mu}{\sigma/\sqrt{n}}. \end{aligned}  The test statisticZ$follows$N(0,1)$distribution. Step 3 Significance Level The significance level is$\alpha = 0.05$. Step 4 Critical Value(s) As the alternative hypothesis is$\text{left-tailed}$, the critical value of$Z\text{is}-1.64$(from Normal Statistical Table). The rejection region (i.e. critical region) is$\text{Z < -1.64}. Step 5 Computation The test statistic under the null hypothesis is  \begin{aligned} Z_{obs}&=\frac{ \overline{x} -\mu_0}{\sigma/\sqrt{n}}\\ &= \frac{8445-8500}{1200/ \sqrt{350 }}\\ &= -0.857 \end{aligned}  Step 6 Decision (Traditional Approach) The test statistic isZ_{obs} =-0.857$which falls$\text{outside}$the critical region, we$\text{fail to reject}$the null hypothesis at$\alpha = 0.05$level of significance. OR Step 6 Decision ($p$-value Approach) This is a$\text{left-tailed}$test, so the p-value is the area to the$\text{left}$of the test statistic ($Z_{obs}=-0.857$) is p-value =$0.1956$. The p-value is$0.1956$which is$\text{greater than}$the significance level of$\alpha = 0.05$, we$\text{fail to reject}$the null hypothesis at$\alpha =0.05$level of significance. $Z$-test for testing mean Example 2 The health of the bear population in Yellowstone National Park is monitored by periodic measurements taken from anesthetized bears. A sample of 54 bears has a mean weight of 182.9 lb. Assuming that is$\sigma$known to be 121.8 lb, use a 0.05 significance level to test the claim that the population mean of all such bear weights is greater than 150 lb. Solution Let$X$denote weight of bear. Given that$n = 54$, sample mean weight of bear is$\overline{x} = 182.9$, the population standard deviation is$\sigma = 121.8$. Step 1 Hypothesis Testing Problem The hypothesis testing problem is$H_0 : \mu = 150$against$H_1 : \mu > 150$($\text{right-tailed}) Step 2 Test Statistic The test statistic for testing above hypothesis testing problem is  \begin{aligned} Z& =\frac{\overline{x} -\mu}{\sigma/\sqrt{n}}. \end{aligned}  The test statisticZ$follows$N(0,1)$distribution. Step 3 Significance Level The significance level is$\alpha = 0.05$. Step 4 Critical Value(s) As the alternative hypothesis is$\text{right-tailed}$, the critical value of$Z\text{is}1.64$(from Normal Statistical Table). The rejection region (i.e. critical region) is$\text{Z > 1.64}. Step 5 Computation The test statistic under the null hypothesis is  \begin{aligned} Z_{obs}&=\frac{ \overline{x} -\mu_0}{\sigma/\sqrt{n}}\\ &= \frac{182.9-150}{121.8/ \sqrt{54 }}\\ &= 1.985 \end{aligned}  Step 6 Decision (Traditional Approach) The test statistic isZ_{obs} =1.985$which falls$\text{inside}$the critical region, we$\text{reject}$the null hypothesis at$\alpha = 0.05$level of significance. OR Step 6 Decision ($p$-value Approach) This is a$\text{right-tailed}$test, so the p-value is the area to the$\text{right}$of the test statistic ($Z_{obs}=1.985$) is p-value =$0.0236$. The p-value is$0.0236$which is$\text{less than}$the significance level of$\alpha = 0.05$, we$\text{reject}$the null hypothesis at$\alpha =0.05$level of significance. $Z$-test for testing mean Example 3 A company that makes cola drinks states that the mean caffeine content per 12 ounce bottle of cola is 40 milligrams. During the test it was found that a random sample of thirty 12-ounce bottles of cola has a mean caffeine content of 38.7 milligrams. The population standard deviation is 7.3 milligrams. At$\alpha = 0.01$test the claim of the company. Solution Let$X$denote caffeine content per 12 ounce bottle of cola. Given that$n = 30$, sample mean caffeine content per 12 ounce bottle of cola is$\overline{x} = 38.7$, the population standard deviation is$\sigma = 7.3$. Step 1 Hypothesis Testing Problem The hypothesis testing problem is$H_0 : \mu = 40$against$H_1 : \mu \neq 40$($\text{two-tailed}) Step 2 Test Statistic The test statistic for testing above hypothesis testing problem is  \begin{aligned} Z& =\frac{\overline{x} -\mu}{\sigma/\sqrt{n}}. \end{aligned}  The test statisticZ$follows$N(0,1)$distribution. Step 3 Significance Level The significance level is$\alpha = 0.01$. Step 4 Critical Value(s) As the alternative hypothesis is$\text{two-tailed}$, the critical value of$Z\text{are}-2.58 \text{ and } 2.58$(from Normal Statistical Table). The rejection region (i.e. critical region) is$\text{Z < -2.58 or Z > 2.58}. Step 5 Computation The test statistic under the null hypothesis is  \begin{aligned} Z_{obs}&=\frac{ \overline{x} -\mu_0}{\sigma/\sqrt{n}}\\ &= \frac{38.7-40}{7.3/ \sqrt{30 }}\\ &= -0.975 \end{aligned}  Step 6 Decision (Traditional Approach) The test statistic isZ_{obs} =-0.975$which falls$\text{outside}$the critical region, we$\text{fail to reject}$the null hypothesis at$\alpha = 0.01$level of significance. OR Step 6 Decision ($p$-value Approach) This is a$\text{two-tailed}$test, so the p-value is the area to the$\text{extreme}$of the test statistic ($Z_{obs}=-0.975$) is p-value =$0.3294$. The p-value is$0.3294$which is$\text{greater than}$the significance level of$\alpha = 0.01$, we$\text{fail to reject}$the null hypothesis at$\alpha =0.01$level of significance. Endnote In this tutorial, you learned the about how to solve numerical examples on$Z$-test for testing population mean. You also learned about the step by step procedure to apply$Z$-test for testing population mean and how to use Z-test calculator for testing population mean to get the value of test statistic, p-value, and z-critical value. To learn more about other hypothesis testing problems, hypothesis testing calculators and step by step procedure, please refer to the following tutorials: Let me know in the comments if you have any questions on$Z\$-test calculator for mean with examples and your thought on this article.

VRCBuzz co-founder and passionate about making every day the greatest day of life. Raju is nerd at heart with a background in Statistics. Raju looks after overseeing day to day operations as well as focusing on strategic planning and growth of VRCBuzz products and services. Raju has more than 25 years of experience in Teaching fields. He gain energy by helping people to reach their goal and motivate to align to their passion. Raju holds a Ph.D. degree in Statistics. Raju loves to spend his leisure time on reading and implementing AI and machine learning concepts using statistical models.