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In this video, we return to the distracted eaters study and compare the average snack

consumption of distracted and non-distracted eaters post-lunch.

As a reminder,

the study was called Playing a Computer Game During Lunch Affects Fullness,

Memory for Lunch, and Later Snack Intake.

The researchers set out to the evaluate the relationship between distraction, and

recall of food consumed, and snacking.

They had a sample of 44 volunteer patients, 22 of them men and

22 of them women, that they randomized into two equally-sized groups.

One group played solitaire on the computer while eating, and

was instructed to win as many games as possible.

In other words, they were asked to focus on the game.

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The other group was asked to eat their lunch without any distractions,

focusing on what they were eating.

Both groups were provided the same amount of lunch and

offered the same amount of biscuits to snack on afterwards.

The researchers measured the snack consumption of subjects in each group.

The study reports average snack consumption levels for

both groups, as well as the standard deviations.

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Suppose we want to estimate how much more, or less,

distracted eaters snack compared to non-distracted eaters.

We would use a confidence interval for

this, which is always of the form, point estimate plus or minus a margin of error.

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So the only new concept here is this new standard error,

which can be calculated as the square root of the sum of the variances for

each group, divided by their respective sample sizes.

Note that we add the two variances even though we're looking for

the standard error of the difference of the two means.

This is not simple to prove mathematically with only the tools we've

learned in this course so far, but conceptually, you can think about it as

bringing together two measures with an inherent variability around them.

These are two sample means.

When you bring two unknowns together, the result should always be more variable,

regardless of whether you're adding them or subtracting them.

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And the degrees of freedom for the t statistic, for comparing two independent

means can be calculated as the minimum of the sample sizes of the two samples,

minus 1.

You might come across a few other estimates for the degrees of freedom for

comparing two means, perhaps in another textbook or so, but what we noted here is

actually not the exact degrees of freedom, which is quite tedious to compute by hand.

The estimate of degrees of freedom we noted here, is a conservative estimate,

since it relies on the lower of the two sample sizes.

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assumption via random sampling or

assignment, and the 10% condition, if you're sampling without replacement.

Note that the 10% condition in this case means that both N1 and

N2 should be less than 10% of their respective populations.

If both of these are met, then we can assume that the observations in our study

are independent of each other with respect to the variable,

the outcome variable that we're studying.

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We also want the two groups to be independent of each other.

Failure to meet this condition is not inherently a problem though, but

it just means that we would need to use methods suited for

dependent, or in other words, paired groups.

We will introduce these methods in another lesson.

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Let's estimate the difference between the average post-meal snack consumption

between those who eat with and without distractions.

The confidence interval will be of the form point estimate, that is

the difference between the two sample means, plus or minus a margin of error.

That is a critical T-score times the standard error of the difference between

the two sample means.

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This is (52.1- 27.1) plus or minus 2.08

x the square root of 45.1 squared over 22 + 26.4 squared over 22,

which yields a standard error estimate of 11.14,

and a margin of error of 23.17 grams,

the confidence interval is then 1.83 to 48.17 grams.

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Next, we'll work on a hypothesis test for evaluating whether these data provide

convincing evidence of a difference between their average post-meal

snack consumption between those who eat with and without distractions.

When doing a hypothesis test, the first step is always to set your hypotheses.

Remember that the null hypothesis always says there's

absolutely nothing going on here.

We could phrase that as the average snack consumption for those who eat with and

without distraction, the difference between those two is 0.

Another way of thinking about this is that the two mus

from the two populations are equal to each other.

Note that I'm using mus and not x-bars because the hypotheses

are always about the populations and never about the samples.

We already know the sample statistics, we don't need to hypothesize about them.

We want to use those sample statistics to say something about

the unknown population parameters.

The alternative hypothesis is then that there is a difference between the two

population means, or that the difference is not 0.

We calculate ofurtest statistic, which is a T-score with 21 degrees of freedom,

as the observed difference, the 25 we saw earlier, minus the null value of 0,

divided by the standard error we calculated earlier of 11.14.

This yields a T statistic of 2.24.

The last step before making a decision on these hypotheses is to find the p-value,

but before we can do that we must sketch the sampling distribution.

We do that and shade the tail areas corresponding to our p-value.

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Let's recap everything we've done so far.

We started with a study where the researchers randomly assigned respondents

into distracted and non-distracted eating groups and

compared their snack intake post-meal.

The sample statistic suggested that the distracted eaters

consumed more snacks on average.

However, just because we observe a difference in the sample means,

doesn't necessarily mean that there is something going on

that is statistically significant in the actual populations.

So, we use statistical inference tools to evaluate if this apparent

relationship between distracted eating and

snacking more provide evidence of a real difference at the population level.

Note that we have a randomized control trial here, so

if we do indeed find a significant result,

we could then talk about a causal relationship between these two variables.

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The confidence interval for

the average difference was 1.83 to 48.17 and the hypothesis test

evaluating a different between the two means yielded a p-value of roughly 4%.

Which means that we would reject the null hypothesis and conclude that

these data do indeed provide convincing evidence that there is a difference

between the average snack intake of distracted and non-distracted eaters.

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And do the results of the confidence interval and the hypothesis test agree?

We used similar methods, so they really should.

We rejected the null hypothesis that set

the difference between the two means equal to 0, and therefore this

null value should not be included in our confidence interval and indeed it's not.

Therefore, the results from the hypothesis test and the confidence interval do agree.