Question

Option 1: Think of a problem that you may be interested in that deals with a comparison of two population means. Propose either a confidence interval or a hypothesis test question that compares these two means. Gather appropriate data and post your problem (without a solution) in the discussion topic. Later, respond to your own post with the solution for others to check their work. For example, you may want to know if the average weight of a rippled potato chip is the same as the average weight of a non-rippled potato chip. You may weigh rippled regular potato chips from a large bag and find weights of 1.7, 1.9, 2.4, 1.3, 1.7, and 2.0 grams. You may weigh non-rippled potato chips from another large bag and find weights of 1.8, 1.6, 1.9, 1.9, and 1.4 grams. Assume a random sample was drawn.

Answer 1

Null Hypothesis

H0: u1 - u2 = 0, where u1 is the mean of first population and u2 the mean of the second.

As above, the null hypothesis tends to be that there is no difference between the means of the two populations; or, more formally, that the difference is zero (so, for example, that there is no difference between the average heights of two populations of males and females).

Equation

$t_independent_means.png$

For rippled potato chips,

M₁: 1.83
SS₁: 0.67
s²₁ = SS₁/(N - 1) = 0.67/(6-1) = 0.13


For non rippled potato chips,

N₂: 5
df₂ = N - 1 = 5 - 1 = 4
M₂: 1.72
SS₂: 0.19
s²₂ = SS₂/(N - 1) = 0.19/(5-1) = 0.05


T-statistic Calculation

s²_p = ((df₁/(df₁ + df₂)) * s²₁) + ((df₂/(df₂ + df₂)) * s²₂) = ((5/9) * 0.13) + ((4/9) * 0.05) = 0.1

s²_M1 = s²_p/N₁ = 0.1/6 = 0.02
s²_M2 = s²_p/N₂ = 0.1/5 = 0.02

t = (M₁ - M₂)/√(s²_M1 + s²_M2) = 0.11/√0.04 = 0.61

The t-value is 0.605. The p-value is .56012. The result is not significant at p < .05.

Also using Excel, data toolpack,

We conclude that, Both populations have Same mean.