This video demonstrates how to solve the following problem:

The daily productivities of machine A and B from different companies are set to be 88 units per day. Machine A is tested for 30 days and the mean is found to be 87.5 and standard deviation is 1.2. While for machine B, it is tested for 35 days, with mean = 88.3 and standard deviation = 1.3. Assume they have the same variances. Can we assume A and B to have the same mean for 5% significant level?

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Always write down the hypothesis before doing any calculations:

H0: MeanA = MeanB

HA: MeanA not equal to Mean B

Then ask the goal: the goal is to disprove H0

Then decide which type of test to use:

1. Machine A and Machine B can be assumed to be independent
2. This is a test of the differences between the means
3. They have the same variances
4. We can assume they are normally distributed.

So, we can use the t-test for the equality of means of 2 populations

t= (MeanA – MeanB) /  (sp^2/n1 + sp^2/n2)

sp^2 = ((n1-1)s1^2 + (n2-1)s2^2 ) / n1+n2-2

Plug in the data, and find that t = 2.56

Then decide the degree of freedom and whether it is one-tail or 2-tail

This is equality H0, so 2-tail

DOF = n1+n2-2 = 63

Look up the t-table with Level of Significant at 5%, then you find that when DOF = 60, it is 2. Therefore, our t-value is 0.56 more than the table.

So, we should reject the hypothesis H0 and therefore, MeanA is not equal to MeanB.