Toasty tortilla company has two different manufacturing plants. company officials want to test whether each plant fills the bags with the same number of ounces. a random sample of tortilla bags from plant a had a mean of 24.2 ounces in each bag. a random sample from plant b had a mean of 23.2 ounces. they randomized the data over 100 trials, and the difference of means for each trial is shown in the dot plot below. what can toasty tortilla company conclude from this study? a dot plot with the values of negative 2.5, negative 2.0, negative 1.5, negative 1.0, negative 0.5, 0.0, 0.5, 1.0, 1.5, 2.0, and 2.5 with frequencies of 3, 6, 9, 12, 0, 9, 8, 5, 1, 0, and 5 respectively.

First of all, future statisticians out there, this is a goofy way to do this study.  We wouldn't usually pick a bag from each pile and record the difference in weight as a trial.  Instead we would weigh a bunch of As, weigh a bunch of Bs, and do a t test to determine is the averages are significantly different.

That said, let's answer the question.

We have n trials, n=3 + 6 + 9 + ... + 5 = 58

The average of the trial differences is a=( 3(-2.5) + 6(-2) + 9(-1.5) + ... + 5(2.5) ) / 58 = 0.379

The variance is the sum of squared deviations from the average divided by n-1=57, and the standard deviation is the square root of that.

σ = 1.359

That's the estimated standard deviation.  What we wish to determine is the probability that we'd get the average different from zero we observed if the true difference was zero. 

We need the standard deviation of the average, which assuming independence blah blah is the standard deviation of a trial divided by the square root of the number of trials:



So we get

t = a/s = 0.379/.178 = 2.1

That's a reasonably big t, and n=58 is big enough we don't need the t tables, we can just use the normal distribution. 

The proper test here is the two side t test, a significant positive or negative difference is what we're looking for.

For a normal distribution 95% of the probability is between plus and minus two sigma from the mean.  See we have t=2.1 so that goes up to about 98%, so a 2% chance of seeing this result by chance.  

We can conclude the weights of the bags from the plants are significantly different.

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