here is how to use the table. ( an excerpt of the table is shown ) one tail0.0500.0250.0100.005 two...
TRANSCRIPT
Estimating When Is Unknown
Student’s - used when the mean and standard deviation are
unknown- we use the sample standard deviation to approximate - the sampling distribution for
With degrees of freedom
Estimating When Is Unknown
Student’s - used when the mean and standard deviation are
unknown- we use the sample standard deviation to approximate - the sampling distribution for
With degrees of freedom
When using a distribution we assume that the distribution is normal.
Estimating When Is Unknown
Student’s - used when the mean and standard deviation are
unknown- we use the sample standard deviation to approximate - the sampling distribution for
With degrees of freedom
When using a distribution we assume that the distribution is normal.It is symmetrical about the mean which is
Estimating When Is Unknown
Student’s - used when the mean and standard deviation are
unknown- we use the sample standard deviation to approximate - the sampling distribution for
With degrees of freedom
When using a distribution we assume that the distribution is normal.It is symmetrical about the mean which is The difference between a distribution and a distribution is the distribution will have thicker tails.
Estimating When Is Unknown
Student’s - used when the mean and standard deviation are
unknown- we use the sample standard deviation to approximate - the sampling distribution for
With degrees of freedom
When using a distribution we assume that the distribution is normal.It is symmetrical about the mean which is The difference between a distribution and a distribution is the distribution will have thicker tails.As the degrees of freedom get larger, the distribution gets closer to a standard normal distribution.
Estimating When Is Unknown
Student’s - used when the mean and standard deviation are
unknown- we use the sample standard deviation to approximate - the sampling distribution for
With degrees of freedom
When using a distribution we assume that the distribution is normal.It is symmetrical about the mean which is The difference between a distribution and a distribution is the distribution will have thicker tails.As the degrees of freedom get larger, the distribution gets closer to a standard normal distribution.
** see me and I will print a copy of the table we will be using from the textbook
Estimating When Is Unknown
Here is how to use the table. ( an excerpt of the table is shown )
One tail 0.050 0.025 0.010 0.005
Two tail 0.100 0.050 0.020 0.010We either have a one tail or two tail test. The numbers represent the area under that tail.
Estimating When Is Unknown
Here is how to use the table. ( an excerpt of the table is shown )
We either have a one tail or two tail test. The numbers represent the area under that tail.
One tail 0.050 0.025 0.010 0.005
Two tail 0.100 0.050 0.020 0.010
d.f. / c …0.900 0.950 0.980 0.990
3
4
- - - - - - - - - - - - - - -
7
8
d.f. is our degrees of freedom = is our confidence interval
Estimating When Is Unknown
Here is how to use the table. ( an excerpt of the table is shown )
We either have a one tail or two tail test. The numbers represent the area under that tail.
One tail 0.050 0.025 0.010 0.005
Two tail 0.100 0.050 0.020 0.010
d.f. / c …0.900 0.950 0.980 0.990
3 2.3533 3.182 4.541 5.841
4 2.132 2.776 3.747 4.604
- - - - - - - - - - - - - - -
7 1.895 2.365 2.998 3.449
8 1.860 2.306 2.896 3.355
d.f. is our degrees of freedom = is our confidence interval
The rest of the table contains the for the corresponding conditions.
Estimating When Is Unknown
Here is how to use the table. ( an excerpt of the table is shown )
We either have a one tail or two tail test. The numbers represent the area under that tail.
One tail 0.050 0.025 0.010 0.005
Two tail 0.100 0.050 0.020 0.010
d.f. / c …0.900 0.950 0.980 0.990
3 2.3533 3.182 4.541 5.841
4 2.132 2.776 3.747 4.604
- - - - - - - - - - - - - - -
7 1.895 2.365 2.998 3.449
8 1.860 2.306 2.896 3.355
d.f. is our degrees of freedom = is our confidence interval
The rest of the table contains the for the corresponding conditions.
EXAMPLE :Find the critical t value for a 0.99 confidence level for a t distribution with sample size
Estimating When Is Unknown
Here is how to use the table. ( an excerpt of the table is shown )
We either have a one tail or two tail test. The numbers represent the area under that tail.
One tail 0.050 0.025 0.010 0.005
Two tail 0.100 0.050 0.020 0.010
d.f. / c …0.900 0.950 0.980 0.990
3 2.3533 3.182 4.541 5.841
4 2.132 2.776 3.747 4.604
- - - - - - - - - - - - - - -
7 1.895 2.365 2.998 3.449
8 1.860 2.306 2.896 3.355
d.f. is our degrees of freedom = is our confidence interval
The rest of the table contains the for the corresponding conditions.
EXAMPLE :Find the critical t value for a 0.99 confidence level for a t distribution with sample size
Solution :
- find the intersection of and
𝑡 0.99=4 .604
Estimating When Is Unknown
Here is how to use the table. ( an excerpt of the table is shown )
One tail 0.050 0.025 0.010 0.005
Two tail 0.100 0.050 0.020 0.010
d.f. / c …0.900 0.950 0.980 0.990
3 2.3533 3.182 4.541 5.841
4 2.132 2.776 3.747 4.604
- - - - - - - - - - - - - - -
7 1.895 2.365 2.998 3.449
8 1.860 2.306 2.896 3.355
EXAMPLE #2:Find the critical t value for a 0.90 confidence level for a t distribution with sample size
Estimating When Is Unknown
Here is how to use the table. ( an excerpt of the table is shown )
One tail 0.050 0.025 0.010 0.005
Two tail 0.100 0.050 0.020 0.010
d.f. / c …0.900 0.950 0.980 0.990
3 2.3533 3.182 4.541 5.841
4 2.132 2.776 3.747 4.604
- - - - - - - - - - - - - - -
7 1.895 2.365 2.998 3.449
8 1.860 2.306 2.896 3.355
EXAMPLE #2:Find the critical t value for a 0.90 confidence level for a t distribution with sample size
SOLUTION :
Find the intersection of and d.f.
Estimating When Is Unknown
Confidence intervals for when is unknown
- the maximal amount of error for a confidence level
Estimating When Is Unknown
Confidence intervals for when is unknown
- the maximal amount of error for a confidence level
Again with some Algebra :
where
Estimating When Is Unknown
EXAMPLE : Find the 99% confidence interval for sample data using student’s t distribution
if the sample data shows , , and
Estimating When Is Unknown
EXAMPLE : Find the 99% confidence interval for sample data using student’s t distribution
if the sample data shows , , and
1. Find the critical t value
Estimating When Is Unknown
EXAMPLE : Find the 99% confidence interval for sample data using student’s t distribution
if the sample data shows , , and
1. Find the critical t valueFor and , ( from table )
Estimating When Is Unknown
EXAMPLE : Find the 99% confidence interval for sample data using student’s t distribution
if the sample data shows , , and
1. Find the critical t valueFor and , ( from table )
2. Find
Estimating When Is Unknown
EXAMPLE : Find the 99% confidence interval for sample data using student’s t distribution
if the sample data shows , , and
1. Find the critical t valueFor and , ( from table )
2. Find
Estimating When Is Unknown
EXAMPLE : Find the 99% confidence interval for sample data using student’s t distribution
if the sample data shows , , and
1. Find the critical t valueFor and , ( from table )
2. Find
3. Find the 99% confidence interval
Estimating When Is Unknown
EXAMPLE : Find the 99% confidence interval for sample data using student’s t distribution
if the sample data shows , , and
1. Find the critical t valueFor and , ( from table )
2. Find
3. Find the 99% confidence interval
Estimating When Is Unknown
EXAMPLE #2 : A company has a new process for manufacturing large artificial sapphires.
In a trial run, 37 sapphires are produced. The distribution of weights is mound
shaped and symmetric. The mean weight for these 37 gems is carats, and the sample standard deviation is carats. Let be themean weight for the distribution of all sapphires produced by
the new process.Find a 95% confidence interval for .
Estimating When Is Unknown
EXAMPLE #2 : A company has a new process for manufacturing large artificial sapphires.
In a trial run, 37 sapphires are produced. The distribution of weights is mound
shaped and symmetric. The mean weight for these 37 gems is carats, and the sample standard deviation is carats. Let be themean weight for the distribution of all sapphires produced by
the new process.Find a 95% confidence interval for .
SOLUTION : ** when there is no column for d.f., use the closest one that is smaller than
Estimating When Is Unknown
EXAMPLE #2 : A company has a new process for manufacturing large artificial sapphires.
In a trial run, 37 sapphires are produced. The distribution of weights is mound
shaped and symmetric. The mean weight for these 37 gems is carats, and the sample standard deviation is carats. Let be themean weight for the distribution of all sapphires produced by
the new process.Find a 95% confidence interval for .
SOLUTION : ** when there is no column for d.f., use the closest one that is smaller than
Therefore
Estimating When Is Unknown
EXAMPLE #2 : A company has a new process for manufacturing large artificial sapphires.
In a trial run, 37 sapphires are produced. The distribution of weights is mound
shaped and symmetric. The mean weight for these 37 gems is carats, and the sample standard deviation is carats. Let be themean weight for the distribution of all sapphires produced by
the new process.Find a 95% confidence interval for .
SOLUTION : ** when there is no column for d.f., use the closest one that is smaller than
Therefore With and , ( from table )
Estimating When Is Unknown
EXAMPLE #2 : A company has a new process for manufacturing large artificial sapphires.
In a trial run, 37 sapphires are produced. The distribution of weights is mound
shaped and symmetric. The mean weight for these 37 gems is carats, and the sample standard deviation is carats. Let be themean weight for the distribution of all sapphires produced by
the new process.Find a 95% confidence interval for .
SOLUTION : ** when there is no column for d.f., use the closest one that is smaller than
Therefore With and , ( from table )
Estimating When Is Unknown
EXAMPLE #2 : A company has a new process for manufacturing large artificial sapphires.
In a trial run, 37 sapphires are produced. The distribution of weights is mound
shaped and symmetric. The mean weight for these 37 gems is carats, and the sample standard deviation is carats. Let be themean weight for the distribution of all sapphires produced by
the new process.Find a 95% confidence interval for .
SOLUTION : ** when there is no column for d.f., use the closest one that is smaller than
Therefore With and , ( from table )
Estimating When Is Unknown
EXAMPLE #2 : A company has a new process for manufacturing large artificial sapphires.
In a trial run, 37 sapphires are produced. The distribution of weights is mound
shaped and symmetric. The mean weight for these 37 gems is carats, and the sample standard deviation is carats. Let be themean weight for the distribution of all sapphires produced by
the new process.Find a 95% confidence interval for .
SOLUTION : ** when there is no column for d.f., use the closest one that is smaller than
Therefore With and , ( from table )
carats carats