Two tail test in Excel | t- Test Hypothesis testing for the mean value (µ) with some examples!!

Hi friends!! This article will help you to understand how to verify or test a claim through t-Test hypothesis. The testing for Two tail test can be conducted if mean of claim is given. Almost all manufacturing company or Service industry uses a Warranty Scheme or product life characteristics on the product to attract the customer for more sell. They also do it to give tough competition to other competitors in the market. If someone wants to claim their warranty scheme it can be done through help of Hypothesis testing on sample data.

How to identify Two tail test?

In above picture it is shown that one tyre company is claiming that average life of tyreis equal to 2 Lakh kilometre so here key word is “average” & Mean value (µ) is 2 Lakh kilometres. If in a claim, we are finding these two key words average & mean value then it is considered as Two tail test

In case of battery manufacturing company, it is claiming that battery is Average back up is equal to 20 hours so in this case key word is “average” & mean value (µ) is 20 hours, it is also the case of two tail test

Let us understand it with some more examples.

  • A soft drink manufacturer claims the average amount in the bottles is 5 liters.
  • The mean amount per sales invoice of a restaurant is Rs. 120

 Solved example: A soft drink manufacturer claims the average amount in the bottles is 5 liters.

 Null Hypothesis Ho: µ = 5.0 liters.

Alternative Hypothesis H1: µ ≠ 5.0 liters.  Two-tail test.

Given Data: – Test the hypothesis for 5% level of significance (α = 0.05, or 5%) with sample size=100.        Degree of freedom (df)= n-1= 100-1= 99. (n= No of sample)

Sample results: Sample size, n=100.    Mean, = 5.05.    Standard deviation (sample) S= 0.065

t-critical values from Excel function for df= 99,

Excel formula – Write in cell box as shown in picture

=t.inv(0.025,99) =   -1.98 for   2.5% area in the left tail.

Similarly other Tstat calculation can be done.

=t.inv(0.975,99) = +1.984 for 97.5% area in the left tail.

  • As tstat value is outside, hence reject the Null hypothesis with 5% risk.
  • Managerial: Average amount in the bottles is not equal to 5 liters

Leave a Comment