Math 119 Chap 8 part 3

Brad Bolton
14 Dec 202016:35
EducationalLearning
32 Likes 10 Comments

TLDRIn this educational video, the presenter concludes Chapter Eight by guiding viewers through a statistical hypothesis test on supermodel heights, using a sample of 10 to determine if their standard deviation significantly differs from the general population's 2.6 inches at a 0.01 significance level. The video demonstrates the process of formulating hypotheses, calculating the test statistic, and determining the p-value, which leads to the rejection of the null hypothesis. Additionally, the presenter discusses confidence intervals, contrasting their use with hypothesis tests, and applies this to leisure time data, concluding with the rejection of the claim that Americans spend an average of 18 hours a week on leisure activities.

Takeaways
  • πŸ“š The script concludes Chapter 8 and introduces the start of Chapter 9, indicating a transition in the educational content.
  • 🧐 A hypothesis test is conducted to evaluate if supermodel heights have a different standard deviation than the general population, using a 0.01 significance level.
  • πŸ“‰ The normal distribution assumption is made for the heights of supermodels, which is crucial for the statistical test being performed.
  • πŸ“ The null hypothesis (H0) states that the standard deviation of supermodel heights equals 2.6 inches, while the alternative hypothesis (H1) claims it is different.
  • πŸ”’ The test statistic is calculated using the formula chi-squared = (n-1) * (sample standard deviation)^2 / (population standard deviation)^2, where n is the sample size.
  • πŸ“Š The sample standard deviation is determined to be approximately 0.8, which is used to compute the chi-squared test statistic.
  • πŸ“ˆ The chi-squared distribution is used to find the p-value, which is compared against the alpha level to make a decision on the null hypothesis.
  • πŸ€” The p-value obtained (5.809 Γ— 10^-4) is less than the alpha level, leading to the rejection of the null hypothesis in favor of the alternative.
  • πŸ“˜ The conclusion drawn is that there is sufficient evidence to suggest that the standard deviation of supermodel heights is different from the general population.
  • πŸ“Š A confidence interval is constructed for the mean time spent on leisure activities by Americans, aiming to test a claim about the average time spent.
  • πŸ€·β€β™‚οΈ The script notes that confidence intervals for proportions cannot be used in the same way as hypothesis tests to make decisions or draw conclusions about proportions.
  • πŸ”‘ The script emphasizes the importance of understanding the process of hypothesis testing and constructing confidence intervals, as well as interpreting the results in the context of the problem.
Q & A
  • What is the significance level used in the hypothesis test for supermodel heights?

    -The significance level used in the hypothesis test for supermodel heights is 0.01.

  • What is the claim being tested in the supermodel height hypothesis test?

    -The claim being tested is that the standard deviation of supermodel heights is different from the standard deviation of the general population's heights, which is 2.6 inches.

  • What is the sample size for the supermodel height hypothesis test?

    -The sample size for the supermodel height hypothesis test is 10 supermodels.

  • What statistical test is used to calculate the test statistic for the supermodel height hypothesis test?

    -The chi-squared test is used to calculate the test statistic for the supermodel height hypothesis test.

  • How is the test statistic calculated in the supermodel height hypothesis test?

    -The test statistic is calculated as (n - 1) times the sample standard deviation squared, divided by the population standard deviation squared.

  • What is the sample standard deviation of the supermodel heights?

    -The sample standard deviation of the supermodel heights is approximately 0.8 inches.

  • What is the result of the chi-squared test statistic calculation?

    -The result of the chi-squared test statistic calculation is approximately 0.852.

  • How is the p-value determined in the hypothesis test?

    -The p-value is determined using the chi-squared distribution and the calculated test statistic, considering the degrees of freedom and the two-tailed nature of the test.

  • What is the p-value obtained from the hypothesis test?

    -The p-value obtained from the hypothesis test is 5.809 times 10 to the negative 4.

  • What decision is made based on the p-value compared to the significance level?

    -Since the p-value is less than the significance level of 0.01, the null hypothesis is rejected, and the alternate hypothesis is accepted, indicating that there is sufficient evidence to support that the standard deviation of supermodel heights is different from the general population.

  • How does the script differentiate between hypothesis testing and constructing a confidence interval?

    -The script explains that while both hypothesis tests and confidence intervals can lead to similar conclusions, the process of constructing a confidence interval is different and is not used to make decisions or draw conclusions for hypothesis tests about proportions.

  • What is the purpose of constructing a confidence interval for the leisure time spent by adults?

    -The purpose of constructing a confidence interval for the leisure time spent by adults is to test whether the amount of time spent on leisure activities is different from the claim that Americans spend an average of 18 hours a week on leisure activities.

  • How does the script suggest determining if the null hypothesis can be rejected based on the confidence interval?

    -The script suggests that if the value stated in the null hypothesis (e.g., 18 hours) is not within the confidence interval, then the null hypothesis can be rejected.

  • What is the conclusion drawn from the confidence interval constructed for the leisure time spent by adults?

    -The conclusion drawn is that there is 90% confidence that the true mean of leisure time is different than 18 hours, as 18 is not within the constructed confidence interval.

Outlines
00:00
πŸ“Š Hypothesis Testing for Supermodel Heights

The script discusses a hypothesis test on a sample of 10 supermodel heights to determine if their standard deviation significantly differs from the general population's standard deviation of 2.6 inches. Using a 0.01 significance level and assuming a normal distribution, the null hypothesis (H0: standard deviation equals 2.6) is tested against the alternative hypothesis (H1: standard deviation not equal to 2.6). The test statistic is calculated using the sample's standard deviation, and a chi-squared distribution is used to find the p-value. The result suggests rejecting the null hypothesis in favor of the alternative, indicating the supermodels' heights have a different standard deviation from the general population.

05:01
πŸ“ˆ Constructing a Confidence Interval for Leisure Time

This paragraph explains how to construct a confidence interval to test the claim that Americans spend an average of 18 hours per week on leisure activities. The researcher uses a sample of 10 adults to calculate the mean and standard deviation, then applies a t-interval calculation on a calculator to find the interval. The confidence level is set at 90%, resulting in an interval from 20.458 to 30.942 hours. The null hypothesis (H0: average leisure time equals 18 hours) is compared to this interval. Since 18 is not within the interval, the null hypothesis is rejected, suggesting that the true mean of leisure time is different from the claimed 18 hours.

10:07
πŸ” Hypothesis Testing vs. Confidence Intervals

The script differentiates between hypothesis testing and confidence intervals, noting that while both can lead to similar conclusions, confidence intervals for proportions cannot be used for hypothesis testing. The example provided involves testing the claim of 18 hours of leisure time spent by Americans, using the same sample data. The paragraph emphasizes the importance of understanding whether the null hypothesis value falls within the confidence interval to make a decision about the hypothesis.

15:08
πŸ“š Conclusion of Chapter Eight and Preview of Chapter Nine

The final paragraph wraps up the discussion on hypothesis testing and confidence intervals, indicating the end of chapter eight. It also provides a preview for the upcoming chapter nine, which will delve deeper into confidence interval information. The speaker intends to create another video to continue the educational content, ensuring that viewers have a clear understanding of the concepts covered.

Mindmap
Decision to Reject or Fail to Reject the Null Hypothesis
Comparison with Claimed Average
Sample Standard Deviation (s)
Sample Mean (xΜ„)
Evidence for Standard Deviation Difference
Rejecting or Accepting the Null Hypothesis
P-Value Interpretation
Two-Tailed Test
Chi-Squared Distribution
Sample Standard Deviation (s)
Chi-Squared Formula
Software for Distribution Functions
Graphing Calculator for Data Analysis
T-Interval Formula
Chi-Squared Formula Application
Guidance for Practical Application
Assumption of Audience's Familiarity
Use of Calculator for Statistical Calculations
Clarification of Terms and Processes
Limitations in Proportions Testing
Applicability to Mean and Standard Deviation
Similarities and Differences
Interpretation of Results
T-Interval Calculation
Sample Data Collection
Confidence Level (e.g., 90%)
Estimating Population Parameters
Decision Making
P-Value Calculation
Test Statistic Calculation
Population Standard Deviation (Οƒ=2.6 inches)
Sample Size (n=10)
Alternative Hypothesis (H1)
Null Hypothesis (H0)
Significance Level (Alpha)
Use of Statistical Tools
Statistical Formulas and Calculations
Engagement with the Audience
Teaching Statistical Concepts
Comparison with Hypothesis Testing
Application in Leisure Time Study
Purpose and Concept
Chi-Squared Test for Standard Deviation
Introduction to Hypothesis Testing
Technical Details
Educational Aspect
Confidence Intervals
Hypothesis Testing
Statistical Analysis of Hypothesis Testing and Confidence Intervals
Alert
Keywords
πŸ’‘Random Sample
A random sample refers to a subset of a population that is selected in such a way that each member of the population has an equal chance of being included. In the video, a random sample of 10 supermodel heights is used to perform statistical tests, which is essential for ensuring that the results are representative of the entire population of supermodel heights.
πŸ’‘Significance Level
The significance level, often denoted by alpha (Ξ±), is the probability of rejecting the null hypothesis when it is actually true. In the script, a 0.01 significance level is used, indicating a very low threshold for rejecting the null hypothesis and thus a very stringent test for statistical significance.
πŸ’‘Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. In the video, the standard deviation of supermodel heights is compared to that of the general population's height to test the claim of a difference. The script mentions a standard deviation of 2.6 inches for the general population.
πŸ’‘Normal Distribution
A normal distribution, also known as a Gaussian distribution, is a type of continuous probability distribution where data points are symmetrically distributed around the mean. The video assumes that the heights of supermodels follow a normal distribution, which is a prerequisite for the statistical tests being discussed.
πŸ’‘Null Hypothesis
The null hypothesis is a statement of no effect or no difference that researchers test to reject in a study. In the context of the video, the null hypothesis is that the standard deviation of supermodel heights is equal to 2.6 inches, which is the standard deviation of the general population's heights.
πŸ’‘Alternate Hypothesis
The alternate hypothesis is a statement that is opposite to the null hypothesis and represents the research hypothesis that the study is designed to support. In the script, the alternate hypothesis is that the standard deviation of supermodel heights is different from 2.6 inches.
πŸ’‘Chi-Squared Test
The chi-squared test is a statistical test used to determine if there is a significant difference between the expected frequencies and the observed frequencies in one or more categories. In the video, the chi-squared test is used to calculate the test statistic for comparing the sample standard deviation to the population standard deviation.
πŸ’‘Test Statistic
A test statistic is a summary of the data used to make a decision in a hypothesis test. In the video, the test statistic is calculated as part of the chi-squared test to determine if there is evidence to reject the null hypothesis regarding the standard deviation of supermodel heights.
πŸ’‘P-Value
The p-value is the probability that the null hypothesis is true given the data. If the p-value is less than the significance level, the null hypothesis is rejected. In the script, the p-value is calculated to be 5.809 x 10^-4, which leads to the rejection of the null hypothesis.
πŸ’‘Confidence Interval
A confidence interval is a range of values, derived from a statistical model, that is likely to contain the value of an unknown parameter. In the video, a confidence interval is constructed to estimate the mean time spent on leisure activities by Americans, which is then used to test the claim of 18 hours per week.
πŸ’‘Hypothesis Test
A hypothesis test is a statistical method used to evaluate the likelihood of obtaining the sample results if the null hypothesis were true. In the video, hypothesis tests are used to determine if there is enough evidence to support the claim that supermodel heights have a different standard deviation than the general population.
Highlights

Introduction to the final section of Chapter Eight with a focus on testing the claim about supermodel heights using a 0.01 significance level.

Explanation of the hypothesis testing process for supermodel heights against the general population's standard deviation.

Assumption that supermodel heights follow a normal distribution for the hypothesis test.

Identification of the null hypothesis (standard deviation equals 2.6 inches) and the alternative hypothesis (standard deviation is not equal to 2.6 inches).

Calculation of the test statistic using the chi-squared formula for a two-tailed test with an alpha of 0.01.

Determination of the sample standard deviation and the calculation of the chi-squared value.

Use of a calculator to find the chi-squared value and the subsequent p-value for the test.

Interpretation of the p-value in the context of the hypothesis test, leading to the rejection of the null hypothesis.

Discussion on the difference between hypothesis testing and constructing a confidence interval for decision-making.

Construction of a confidence interval for the average leisure time of Americans based on a sample.

Use of a calculator to perform a t-interval calculation for the leisure time data.

Comparison of the null hypothesis value (18 hours) with the confidence interval to decide whether to reject the null hypothesis.

Conclusion that there is sufficient evidence to reject the claim that the average leisure time equals 18 hours.

Explanation of how the confidence interval can lead to a similar conclusion as the hypothesis test in this scenario.

Discussion on the potential outcomes if the null hypothesis were set at a different value within the confidence interval.

Announcement of the completion of Chapter Eight and the upcoming video for Chapter Nine focusing on more confidence interval information.

Transcripts
00:03

okay let's finish up chapter eight so this will  be the last few pages from chapter eight um and  

00:08

then I'll try to post another video for the start  of chapter nine so we have this problem it says  

00:14

consider a random sample of 10 supermodel heights  listed below we wish to use a 0.01 significance  

00:19

level um to test the claim that supermodel heights  are different than standard deviation of heights  

00:24

in general population which is 2.6 inches assume  that heights of supermodels follows a normal  

00:29

distribution so we got our normal distribution  we have our sample size of n right 10 supermodels  

00:37

um we have our alphas right here um and again  I'm just kind of going through the motions here  

00:42

clicking on every or checking everything that  I do actually know for this particular problem  

00:47

okay so um super model amount of different standard  deviation and they say claim here's the claim that  

00:56

the super model height or heights are different  than the standard deviation of the heights of  

01:00

the population which is 2.6 so that sentence  right there gives us our claim and our alternate  

01:07

hypothesis so we're going to start with that so  our claim and our alternate or opposite claim

01:18

remember these are both claims and this is  how we can write notation for this problem  

01:23

so we have that the claim is that the standard  deviation is different not equal to 2.6 which  

01:34

is the general population and that the  standard deviation is equal to 2.6 okay so

01:44

we're working with this let's see  what we got so now we need our null  

01:48

okay and we have our null to be  standard deviation equals 2.6  

01:54

and the alternate hypothesis is the standard  deviation is not equal to 2.6 so once again  

02:02

we know that this is a two-tail test  and our alpha is worth .01

02:12

okay and again we've repeated this many times  hopefully you guys are getting the hang of it  

02:16

the alternate hypothesis tells us what type  of tail test it's gonna be so we're going  

02:22

to calculate our test statistic I'll do it  right here chi squared is equal to n minus 1

02:34

times the standard deviation squared all over  the standard deviation squared of the population  

02:39

so what we need to do is we need to know that n  minus 1 is 9 but we need to take our calculator  

02:47

and plug the list of these values into our  calculator so we can create our list so we're  

02:53

going to create our list you guys are going to go  and you know go to um stat edit list one and then  

02:59

plug these values into list one and then you're  gonna do stat and then single variable calculation  

03:05

okay and you will get your standard deviation  I'll write it up here or I'm sorry x bar is equal

03:14

to 69.825 and our s our standard  deviation of the sample is worth  

03:23

0.7997 so I'm going to say 0.78 or I'm sorry  0.800 or 0.8 okay so let's just clean this up oops  

03:32

we're going to do the standard deviation of times  0.8 squared divided by um let's see here 2.6

03:44

squared I guess I don't need the parentheses

03:48

okay so then when I hit enter on my calculator  I'm going to do 9 times 0.8 squared divided by  

03:57

2.6 squared I get 0.852 um I guess  .8521 if you want to round off to

04:07

four decimal places we could use  three but it four should be fine  

04:12

so I get .8521 okay well you know  what let's round off to three because  

04:18

normally they don't go much more than three  on chi squared suppose so let's just go 0.852  

04:24

um now we need to calculate our p value okay so  once again we can do that with our calculator

04:31

so we're going to draw our chi-squared cdf uh  distribution our chi squared distribution i'm  

04:37

sorry and we know that 0.852 is somewhere  right there and it's a two-tailed test so I  

04:45

have this area and then I know somewhere over here  there's a matching amount of area now it'll have a  

04:50

different shape but it'll be the same amount of  green if you if you had the same amount of area  

04:56

um this one will on the right hand side it'll  be thinner the one on the left hand side will  

05:00

be a bigger slice but they both will be the same  amount of area okay so we don't know this number  

05:07

but we don't need to know because what we're  looking for is the p-value and the p-value equals  

05:13

and we're going to go into second vars on my  calculator it's number eight i'm going to go into  

05:20

chi-squared cdf and let's see what they ask for  they want to know the lower bound so we can say  

05:27

zero because we know that standard deviation  starts at zero right this is a zero right here  

05:33

and then we can go to our upper bound of 0.852 and  degrees of freedom of nine so we're gonna go zero  

05:40

.852 comma nine degrees of  freedom and let's see what we get for our p value

05:48

I get 2.904 times 10 to the negative 4 

05:55

but that's actually just the p value for that  isn't there also the matching p value oh I'm sorry  

06:01

that's the p value for way over here  on the other side but we also have  

06:04

this matching p value right there so we  have to double this number so we double

06:13

to get our p value so this is actually

06:18

times two so I get 5.808 and  I guess 09 times 10 to the negative 4

06:30

so this is my true p value the other one  this number here is just the left-hand  

06:34

side of my p-value but I need the whole  p-value because this is a two-tailed test  

06:39

so it's 5.809 times 10 to the negative  4 but it's still going to be the same  

06:44

result because they're both very small  numbers so we say since our p-value

06:54

5.809 times 10 to the negative 4 is less than

07:04

our

07:08

alpha equals 0.01 we reject the null and accept  the alternate hypothesis okay and then you guys  

07:28

I think you're getting used to doing interpret  your decision in the context of the problem  

07:32

we say there is sufficient evidence to support  explain that the standard deviation of supermodel

08:02

supermodel

08:05

heights is different

08:17

from the general population of 2.6  okay that's what our decision was okay

08:31

so now we're also going to use a

08:36

a confidence interval and a hypothesis s um  kind of in the same section so decision made  

08:44

on a conclusion drawn from hypothesis tests  about the mean and the standard deviation  

08:47

can be made from constructing an  appropriate confidence interval  

08:50

the same cannot be said about confidence intervals  for proportions being used to make decisions and  

08:54

draw conclusions for hypothesis tests for the  proportions okay so let's take a look at this

09:06

so it says a past study claims that  adults americans spend too much time  

09:11

on average of 18 or americans spend an  average of 18 hours a week on leisure  

09:16

activities a researcher wanted a researcher  wanted to test this claim she took a random  

09:21

sample of 10 adults and asked them about the  time they spend per week on leisure activities  

09:26

their responses and hours as follows  so we've already done this problem okay

09:33

but now we're going to assume that now we're going  to see let me see this one assume the time spent  

09:39

on leisure activities by all students are normally  just reuse the sample of information to construct  

09:43

a confidence interval the confidence interval is  different than a hypothesis test but oftentimes  

09:48

they will lead to similar conclusions  the hypothesis test had an alpha of 0.10  

09:53

to test whether the amount of time spent on  leisure is different than what has been claimed  

09:59

so here's here's kind of the thing  we're going to deal with here so

10:07

assume time spent on leisure activities by  adults is normally distributed use the sample  

10:11

information to construct a confidence  interval so we're going to construct a  

10:15

confidence interval for this problem so you guys  remember how to do a confidence interval okay

10:23

confidence interval okay for t was we could  use oh well we could use our calculator we  

10:30

could use t interval that was the  easier way that I showed you guys  

10:36

so we could do a t interval and then  let's build our confidence interval

10:43

so we're going to have to plug this into  our calculator so let's go into stat  

10:49

and edit and let's plug this in  14 25 22 38 16 26 19 23 41 33  

11:03

so we have one two three four five six  seven eight nine ten so I think we're good  

11:08

so now I'm gonna go into stat test  and then we're gonna do t interval  

11:16

on my calculator it's number eight and this  time we're going to use data because we  

11:19

just plugged it in and our confidence level is  what um it was 10 percent so we're going to say 90 percent 0.9

11:33

okay so I get oops um let's  go into my interval would be  

11:40

20 point I guess 458 and 30.942 okay so now it says assume  

11:52

that the time spent on leisure activities  and by all adults are normally distributed  

11:56

so we got this we're doing good use the sample  information to construct a confidence interval  

12:00

okay to test whether the amount of time spent on  leisure is different than what has been claimed  

12:06

okay so the claim was what the null was that  mu was equal to 18 and the alternate was that  

12:16

mu is not equal to 18. okay so here's the thing  we're trying to see if 18 is in there or not okay

12:32

so if 18 is in there then we cannot reject the  null okay but if 18 is outside of this interval  

12:41

then we can reject the null okay so let's see here  if we had a number line right and we had 20.458

12:52

and then over here is 30.942 so we know that  somewhere in here is the true mean we're not  

12:59

exactly sure what it is but the true mean is  somewhere in between 20 and 30 about 31. the  

13:06

null is about 18 or the null is 18. well  that's over here somewhere so here's 18.  

13:15

okay it is not in this region right here  it's not inside the interval so that means  

13:21

that we can reject the null 18 is  outside so we reject the null so since

13:30

18 is not in our interval

13:37

we reject

13:43

the claim that

13:51

or where we could say you know what let's say  it this way this might be a little bit better

13:57

we reject the leisure time

14:07

equaling 18.

14:13

um so we are so we could say we are 90 percent confident that the true mean of leisure time

14:36

oops

14:42

is different than

14:49

18. but I want you guys to understand what we kind  of did is didn't we reject the null and accept the  

14:55

alternate hypothesis so it's another way to think  about it right so because the null hypothesis  

15:02

was not inside the confidence interval  it's a way of being in outside of our  

15:07

um it's not inside of what we would accept so  this is a rejectable null okay now what if 18 was  

15:16

so instead of saying that null was 18 what if  they had said the null was like 23 well you  

15:21

don't know for sure but is it possible that 23 is  true that it could be the true mean time yes it  

15:27

is possible so then we would not be able to reject  the null we would have to fail to reject the null  

15:32

okay so let's look at this other problem  let's see how much more we've got left okay

15:50

okay so um this one I don't think I think we  already did something very similar to this  

15:59

so um I think I'm gonna have to write this  one up differently because I think there's  

16:04

an error in it I'm gonna have to double check  it for you guys um so I'm gonna go ahead and  

16:08

write this one up separately on a different um  problem I don't believe we did this one already  

16:14

so I'd like you guys to go over these two okay  and this should be the end of chapter eight  

16:21

okay um and then I will have another video  on chapter nine and we're going to go over  

16:26

more confidence interval information  okay that'll do it for this quick video