A one tail is appropriate if the estimated value may depart from the reference value in only one direction.

** Example **: Whether machine is producing more than two percent defective products or less than two percent defective products.

There are two tests in one sample test. One is Lower tail test that determine whether the mean is greater than the reference value and other one is Lower tail test that determines whether mean is lesser than the reference value.

Collecting the summary of above explanation, we may display Lower tail and Upper tail test in below bell shaped curve.

One Tail Test vs Two Tail Test :

A two-tailed test is appropriate if the estimated value may not be equal to the reference value and a one-tailed test is appropriate if the estimated value may depart from the reference value in only one direction .

**Two Tail Problem** : A perfume manufacturing company producing 100cc of with standard deviation of 4cc. There are 200 bottles randomly picked and average was found to be 104 cc. Has mean value changed?

Here Null Hypothesis(H0): µ=100

Alternate Hypothesis(Ha): µ≠100

**One Tail Problem:** A perfume manufacturing company producing 100cc of with standard deviation of 4cc. There are 200 bottles randomly picked and average was found to be 104 cc. Has mean value increased?

Here Null Hypothesis(H0): µ<=100

Alternate Hypothesis(Ha): µ>100

We got to know about one and two tail tests and now let’s talk about One Sample z,t and p test. Later we will see two sample z,t, and p test.

One Sample Z test:

It is one of the most simple hypothesis test and use below formula to calculate the z statistics:

We need to define the null hypothesis and alternate hypothesis. Then calculate the z statistics using the formula and compare with the critical value of z and interpret the result whether null hypothesis is rejected or failed to reject the null hypothesis.

Let’s solve the problem related to Z statistics:

**Problem-**A perfume manufacturing company producing 100cc of with standard deviation of 4cc. There are 100 bottles randomly picked and average was found to be 104 cc. Has mean value changed? (95% confidence level)

**Step-1:**

Null Hypothesis(H0): µ=100

Alternate Hypothesis(Ha): µ≠100

**Step-2:**

x=104,µ=100,n=100,σ=4

Z=(104-100)/4/sqrt(100) =10

**Step-3: **For α=.05(95% confidence level) Z critical(two tail test) =1.96

**Step-4: **Z *calculated > Z critical, *means ** rejected the null hypothesis**.We can conclude that mean has been changed for the population.

**One Sample T test**

T test is used when there are less number of samples (<30) taken from the population and standard deviation of sample was given. T test can be calculated from below formula:

As we have already seen that state the null hypothesis and alternate hypothesis. Then now calculate the t statistics using the formula and compare with the critical value of t and interpret the result whether null hypothesis is rejected or failed to reject the null hypothesis.

Let’s solve the problem related to T statistics:

**Problem-**A perfume manufacturing company producing 100cc . There are 9 bottles randomly picked and average was found to be 101 cc and standard deviation of sample is 3. Has mean value changed?(95% confidence level)

**Step-1:**

Null Hypothesis(H0): µ=100

Alternate Hypothesis(Ha): µ≠100

**Step-2:**

x=101,µ=100,n=9,s=3

t=(101-100)/3/sqrt(9) =1

**Step-3: **For α=.05(95% confidence level) t critical(two tail test)=3.18

**Step-4: **Z *calculated < Z critical, *means ** failed to reject the null hypothesis**.We can conclude that mean has not been changed for the population.

**t and z statistics criteria : **

Lets understand the p value and one sample p test in the next article.

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