HT (for the mean) examples

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Del curso dictado por Duke University

Inferential Statistics

1265 calificaciones

Duke University

Inferential Statistics

1265 calificaciones

Curso 2 de 5 en SpecializationStatistics with R

This course covers commonly used statistical inference methods for numerical and categorical data. You will learn how to set up and perform hypothesis tests, interpret p-values, and report the results of your analysis in a way that is interpretable for clients or the public. Using numerous data examples, you will learn to report estimates of quantities in a way that expresses the uncertainty of the quantity of interest. You will be guided through installing and using R and RStudio (free statistical software), and will use this software for lab exercises and a final project. The course introduces practical tools for performing data analysis and explores the fundamental concepts necessary to interpret and report results for both categorical and numerical data

De la lección

Inference and Significance

Welcome to Week Two! This week we will discuss formal hypothesis testing and relate testing procedures back to estimation via confidence intervals. These topics will be introduced within the context of working with a population mean, however we will also give you a brief peek at what's to come in the next two weeks by discussing how the methods we're learning can be extended to other estimators. We will also discuss crucial considerations like decision errors and statistical vs. practical significance. The labs for this week will illustrate concepts of sampling distributions and confidence levels.

Another Introduction to Inference4:11

Hypothesis Testing (for a mean)14:02

HT (for the mean) examples9:20

Conoce a los instructores

Mine Çetinkaya-Rundel
Associate Professor of the Practice
Department of Statistical Science

[MUSIC]

Researchers investigating characteristics of gifted children collected data

from schools in a large city on a random

sample of 36 children who were identified as gifted

children soon after they reached the age of four.

In this study, along with variables on the children,

the researchers also collected data on their mothers' IQ scores.

The histogram shows the distribution of these data, and also provided our some

sample statistics.

We're asked to perform a hypothesis test to evaluate if

these data provide convincing evidence of a difference between the average

IQ score of mothers of gifted children And the average IQ

score for the population at large, which happens to be 100.

We're also asked to use a significance level of .01.

First, we set the hypothesis.

Before we do that, though, let's define our parameter of interest.

Which is a mean in this case.

So, we are going to let mu equal

the average IQ score of mothers of gifted children.

Therefore, the no hypothesis is that mu is equal to 100

and the alternate hypothesis is that mu is different than 100.

Next we're going to calculate the point estimate.

Which is already given to us as the sample means.

So our x bar is simply a 118.2.

That is the average IQ score of these 36 mothers.

Then we check the conditions.

Note that we're not specifically told to do so.

But we always should check the conditions anyway.

The first condition is about independence.

We have a random sample, we are told, and

36 is definitely less than 10% of all gifted children.

Therefore we can assume independence, but let's get into a little

bit of detail as to what do we mean by, we can assume independence?

What we mean here is that we can assume that the I, IQ score.

Of one mother in this sample, is independent of another one.

The other condition is about the sample size and skew.

We know that the sample size is greater than 30

and looking at the histogram, the sample is not skewed.

Therefore, we can

expect to see a nearly normal sampling distribution of the sample mean.

Then, x bar, our sample mean, is distributed nearly normally with mean

equal to 100 the null value and standard area equal to s over square root of n.

S happens to be 6.5 here divided by square root of 36.

We get a standard error of 1.083.

So now we know the shape, the center, and the spread of our sampling distribution.

Which is going to come in handy when we're trying to

draw the sampling distribution, shade the p-value, and calculate the test statistic.

We have a nearly normal distribution, centered at 100.

And since the p-value is defined as the probability

of observed or more extreme outcome Given the null hypothesis

is true, we need to shape the tails of the curve beyond our observed mean of 118.2.

So, we want to shade here at the higher end.

However, since the alternative hypothesis is two-sided, we also need

to shade the symmetric tail area on the other side.

The cutoff for the lower tail will be equidistant

to the mean as the cutoff for the upper tail.

And since 118 minus 100 is 18.2, we subtract 18.2 from

100, and the lower lower tail, find the lower tail cutoff to be 81.8.

And we want to shade everything below that as well.

We can already see that these tail areas will be tiny.

But let's formally solve for the p-value for completeness anyway.

First,

we calculate a z score as the observed sample mean 118.2 minus the null

value 100 divided by the standard error. 1.083.

Our z-score comes out to be 16.8.

Such a high z-score means that the

observation is unusually far from the center, and

hence, the p-value, which is going to be the sum of the two tail areas here,

will be approximately 0.

Lastly we need to make a decision and interpret it in context of the data.

This smaller the p value, the stronger the evidence against the null.

And since we have a tiny p value, we have very strong evidence against the

null hypothesis, therefore, we reject the null hypothesis

and conclude that the data provide convincing evidence.

Of a difference between the average IQ score of mothers of

gifted children and the average IQ score for the population at large.

Let's try our hand at another question.

A statistics student interested in sleep habits of domestic cats

took a random sample of 144 cats and monitored their sleep.

The cats slept an average of 16 hours per day.

According to our online

resources, domestic dogs actually sleep on average 14 hours a day.

We want to find out if these data provide convincing evidence of different

sleeping habits for domestic cats and dogs with respect to how much they sleep.

Note that the test statistic calculated was 1.73.

So, we're told our sample mean

as 16 hours.

We are also told that dog sleep on average 14 hours per day.

And therefore our null hypothesis is going to be that mu is equal to 14, and here

note that mu actually speaks to the average number of hours cats sleep.

So that's our parameter of interest.

And our alternative hypothesis is that mu is different than 14.

Now that we have our hypothesis set.

We can now draw our curve, mark our test statistic

and shade the tail areas corresponding to the p value.

Using an Applet or, or the table, we can find that the lower tail area

comes out to 0.0418 for a z score of 1.73.

Because the distribution is symmetric, the upper tail area is also going

to be point 0.0418 and therefore the total p value for this hypothesis test is simply

going to be one of the tail areas times 2, which comes out to be point 0.836.

Next question is what is the interpretation

of this p-value in context of these data?

We know that the p-value is basically the probability of observed or more

extreme outcome given the nal hypothesis is true.

Let's also bring back some of our sample statistics that we've seen

earlier, as well as our hypothesis to help us construct a phrase.

That actually interprets the p value in context of these data.

First, let's focus on what we mean by observed.

The observed outcome is x bar of 16 hours

per day.

So, the observed outcome is cats sleeping on average 16

hours per day, from a random sample of 144 cats.

Something more extreme could actually be in the

positive direction, so 16 or greater, or you

could also, remember, be in the negative direction

as well, because our alternative hypothesis is two-sided.

And to figure out where the cutoff value for

the negative side of the distribution is going to be.

What we want to do, is we want to venture equal distance

from the mean, as we did between 14 and 16.

That's two hour difference. So we would do 14 minus 2.

12 hours or lower is also of interest to us.

Lastly, the no hypothesis being true.

Simply means that mu is equal to 14 or in other words that

truthfully, cats do sleep 14 hours on average.

So, putting all of this together, this is the

probability of obtaining a random sample of 144 cats.

That sleep 16 hours or more or 12 hours or less, on

average, if in fact cats truly slept 14 hours per day on average.

And we saw that the probability of this happening

is 0.0836, so roughly 8%.

According to some of the guidelines we've been using

so far, that would actually be considered kind of high.

So once again, to parse through this

longish phrase that we've come up with, the

observed or more extreme outcome is simply the obtaining a random sample of 144 cats.

That sleep either 16 hours or more

per day or 12 hours per less or less per day and the part of it that's about the

null hypothesis being true is basically saying, if in fact

that cats truly slept 14 hours per day, on average.

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