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Testing if Two Means are Different

Note:  This is a brief summary of part of Section 11.2.  There is much more information in the section which you are invited to read.

The problem:  You have two samples from independent populations, that is, thee data in one sample does not affect data in the other sample.  The sample means are and  the sample standard deviations are s₁ and s₂ and the sample sizes are n₁ and n₂.  You want to test if the population means m₁ and m₂  are different.

Note:  When setting up these tests start with the alternative hypothesis, which is what you are trying to show to be true or false.  In a two tailed test it is . (Alternatively m₁ - m₂ ≠ 0)  For a left-tailed test it is  (Alternatively m₁ - m₂ < 0) and for a right-tailed test it is (Alternatively m₁ - m₂ > 0)  (At least in any example I give you)

A Two-Tailed Test

Gardener Joe wants to try a new, and cheaper, fertilizer on his broad beans.  He fertilizes 30 plants with the old fertilizer and finds that the mean yield is 31 ounces with a standard deviation of 5 ounces.  He fertilizes 15 plants with the new fertilizer and finds the mean yield is 29 ounces with a standard deviation of 6 ounces.  Can he conclude that the mean yields are different.  Use a = 0.05.

The data can be summarized as follows.  It does not matter which data set uses the subscript 1.

How to set it up:                                H₀: m₁ = m_{2                 Alternatively}      H₀: m₁ - m₂ = 0

                                                         H₁: m₁ ≠ m₂                                                H₁: m₁ - m₂ ≠ 0

                                                         Reject H₀ if P < 0.05

Since t is in the right tail compute the area right of t and double it because the test is two tailed

P = 2tcdf(ans, 10000, 14) = 0.2848 Note:  d.f. = 15- 1

Do not reject H₀ because P > a.  The new fertilizer does not make a difference.

A Right-Tailed Test

Change the assumption population standard deviation is the sample standard deviation

The data can be summarized as follows

Reject H₀  if P < 0.01

Compute =1.868

Compute P = tcdf(ans, 10000, 77) = 0.328

Because P is larger than the level of significance do not reject the null hypothesis.  The prep course does not improve the scores.

Note:  Whenever you are wanting to show the test is right tailed.  You will always get a positive value for t and you compute the area right of t to find P.

A Left-Tailed Test

Change the assumption to assume the population variances are unequal

The data can be summarized as follows

Reject H₀  if P < 0.05

Compute

Compute P = tcdf( −10000, ans,11) = 0.0035

Because P is smaller than the level of significance reject the null hypothesis.  Students in traditional lab courses scored lower

Note:  Whenever you want to show the test is left tailed.  You will always get a negative value for t and you compute the area left of t to find P.