Statistica - esercizi

A help to make inference about a population variance is to use the sampling distribution of :

2

σ

this distribution has a chi-square distribution whenever a simple random sample of size n is selected from a

normal population, so the notation is:

2

s

(n−1) 2

χ distribution

2

σ

We can use the chi-square distribution to develop interval estimates and conduct hypothesis tests about a

population variance.

Interval Estimation



We cannot estimate the exact value of the variance, so we need to compute an interval estimation. Intervals

are commonly chosen such that the parameter falls within with a 95 or 99 percent probability, called

confidence intervals

confidence coefficient. Hence, the intervals are called ; the end points of such an

interval are called upper and lower confidence limits and containing a population parameter that is

established by calculating statistics from values measured on a random sample taken from the population

probability theory

and by applying the with a fidelity that represent those of the entire population. The

probability tells what percentage of the time the assignment of the interval will be correct but not what the

chances are that it is true for any given sample. Of the intervals computed from many samples, a certain

percentage will contain the true value of the parameter being sought.

So we can define the interval estimation as:

− −

2 2

( n 1

) s ( n 1

) s

σ

≤ ≤

2

χ χ

2 2

α α

−

/ 2 (

1 / 2 )

Proof 2

n−1 s

( )

2 2

≤ ≤❑

❑

1−α α

/2 /2

2

σ

Working on a leftmost quantity: 2

2

n−1 s

( )

2 ≤

❑

1−α /2 2

σ

2 2 2

σ ≤ n−1 s

( )

❑

= 1−α 2

/ 2

n−1 s

( )

2

σ ≤

= 2

❑

1−α /2

For the rightmost quantity the procedure is similar, so we obtain:

2

n−1 s

( ) 2

≤σ

2

❑

α /2

The result of the combination of two expression is:

− −

2 2

( n 1

) s ( n 1

) s

σ

≤ ≤

2

χ χ

2 2

α α

−

/ 2 (

1 / 2 )

Hypothesis Testing

Hypothesis testing is a method for testing a claim or hypothesis about a parameter in a population, using

data measured in a sample. In this method, we test some hypothesis by determining the likelihood that a

sample statistic could have been selected, if the hypothesis regarding the population parameter were true.

We can use three different types of hypothesis testing:

CASE 1 (Left Tailed Test)

σ σ

≤

2 2

H :

0 0 Hypotheses

σ σ

>

2 2

H :

a 0

− 2

( n 1

) s

χ =

2 σ 2

0 Test Statistic 3

χ χ

>

2 2

α Rejection Rule 2

❑

Reject H if (where is based on a chi-square distribution with n - 1 degrees of freedom)

α

0

or α

Reject H if p-value <

0

CASE 2 (Right Tailed Test)

σ σ

≥

2 2

H :

0 0 Hypotheses

σ σ

<

2 2

H :

a 0

− 2

( n 1

) s

χ =

2 σ 2

0 Test Statistic

χ χ

<

2 2 α

−

(

1 ) Rejection Rule 2

2 2 ❑

❑ >❑

Reject H if or (where is based on a chi-square distribution with n

α (1−α )

/2

0 α

- 1 degrees of freedom) or Reject H if p-value <

0

CASE 3 (Two Tailed Test)

σ σ

=

2 2

H :

0 0

Hypotheses

σ σ

≠

2 2

H :

a 0

− 2

( n 1

) s

χ =

2 σ 2

0 Test Statistic 4

χ χ

<

2 2 α

−

(

1 / 2 ) Rejection Rule 2 2

2 2 ❑ ∧❑

❑ >❑

Reject H if or (where are based on a chi-square

2 α

α (1−α )/ /2

/2

0 α

distribution with n - 1 degrees of freedom) or Reject H if p-value <

0

So, we can define now the hypothesis tests about a population variance as follows:

2

n−1 s

( )

2

❑ = 2

σ 0

2

χ

Where has a chi-square distribution with n-1 degrees of freedom

As we’ve seen previously,two tools can be used to check the hypothesis testing:

p-value: p-value estimate the strength of rejection of null hypothesis (H0) when that hypothesis is

 true. The lower p-value, greater the strength of rejection;

critical value approach: involves to determine "likely" or "unlikely" by determining whether or not the

 observed test statistic is more extreme than would be expected if the null hypothesis were true. If the

test statistic is more extreme than the critical value, then the null hypothesis is rejected in favor of the

alternative hypothesis. If the test statistic is not as extreme as the critical value, then the null

hypothesis is not rejected.

Inferences About Two Population Variances

Sometimes, we need to compare the variances concerning to different production process, methods, etc. so

we will be using the random sample data which are independent (one from population 1 and the other from

2 2

s s

population 2). The value of and will be necessary to make inference about the variance of two

1 2

2 2

σ σ

populations, indicated with and respectivly. Every time the two variances are equal, the

1 2

2 2

s s

sampling distribution of the ratio / is defined as follow:

1 2

2

s 1 2 2

n n σ =σ

1 2

with size equal to for the population 1 and for the population 2 and 1 2

2

s 2

In this case:

2 n

s

• 1

is the sample variance for the random sample of from population 1

1

2 n

s

• 2

is the sample variance for the random sample of from population 2

2

n n

−1 −1

With degrees of freedom for the numerator and degrees of freedom for the denominator.

1 2

5

We know that the F distribution isn’t symmetric so the value for this distribution cannot be negative. It’s

interesting to observe how the shape of any particular F distribution varies when the degrees of freedom for

the numerator and the denominator change.

Hypothesis Testing

ONE-TAILED TEST

σ σ

≤

2 2

H :

0 1 2 Hypotheses

σ σ

>

2 2

H :

a 1 2

2

s

= 1

F 2

s 2 Test Statistic

Rejection Rule F F n

F −1

α α 1

Reject H if > the value of is based on an F distribution with (numerator) and

0

n −1

2 (denominator) degrees of freedom.

TWO-TAILED TEST

σ σ

=

2 2

H :

0 1 2 Hypotheses

σ σ

≠

2 2

H :

a 1 2

2

s

= 1

F 2

s 2 Test Statistic

Rejection Rule F F n

F −1

α α 1

Reject H if > the value of is based on an F distribution with (numerator)

/2 /2

0

n −1

2

and (denominator) degrees of freedom. 6 21 22

TEST STATISTIC FOR HYPOTHESIS TESTS ABOUT POPULATION VARIANCES WITH σ = σ

2

s

1

F= 2

s

2

Denoting the population with the larger sample variance as the population 1, the F distribution for test

n n

−1 −1

1 2

statistics has degrees of freedom for the numerator and degrees of freedom for the

denominator.

If we did not construct the test statistic in this way, the lower tail areas of probabilities would be needed with

additional calculations or more extensive F distribution tables would be required.

EXERCISES

Interval Estimation

• The variance in drug weights is critical in the pharmaceutical industry. For a specific drug, with weights

2

measured in grams, a sample of 18 units provided a sample variance of s = 0.36.

a) Construct a 90% confidence interval estimate of the population variance for the weight of this drug.

n = 18

2

s = 0.36

α = 0.10 2

n−1)s

(

2

χ = 2

σ 2

n−1 s

( )

2 2

2 2 2 ≤ ≤❑

≤ χ ≤❑ ❑

❑

= = 0.95 0.05

0.95 α /0.05 2

σ

2 2 2 2

σ ≤ n−1 s ≤❑

( )

❑

= 0.095 0.05

− −

2 2

( n 1

) s ( n 1

) s

σ

≤ ≤

2

0

. 05 0

.

095

= 17∗0.36 17∗0.36

2

σ

= ≤ ≤

27.587 8.672

2

σ

= 0.222 ≤ ≤ 0.706

We are sure at 90% that the value about weight of this drug is between 0.222 and 0.706 grams.

b) Construct a 90% confidence interval estimate of the population standard deviation.

7

2

σ

= √0.222 ≤ √ ≤ √ 0.706

σ

= 0.471 ≤ ≤ 0.840

We are sure at 90% that the value about weight of this drug is between 0.471 and 0.840 grams.

• A daily car rental rates for a simple of eight cities has a mean equal to 44.25, a variance equal to 31.152

and a standard deviation equal to 5.582.

a) What is the 95% confidence interval estimate of the variance of car rental rates for the population ?

n = 8

x

́ =44.25

2

s = 31.152

α = 0.10 2

n−1)s

(

2

χ = 2

σ 2

n−1 s

( )

2 2

2 2 2 ≤ ≤

≤ χ ≤❑ ❑ ❑

❑

= = 0.975 0.025

0.975 α 0.025

/ 2

σ

2 2 2 2

σ ≤ n−1 s ≤❑

( )

❑

= 0.0975 0.025

− −

2 2

( n 1

) s ( n 1

) s

σ

≤ ≤

2

0

. 025 0 . 0975

= 7∗31.152 7∗31.152

2

σ

= ≤ ≤

16.013 1.690

2

σ

= 13.618 ≤ ≤ 129.032

We are sure at 95% that the value for daily car rental rates is between 13.618 and 129.032.

b) Construct a 90% confidence interval estimate of the population standard deviation.

2

σ

= √13.618 ≤ √ ≤ √ 129.032

σ

= 3.690 ≤ ≤ 11.359

We are sure at 95% that the value for daily car rental rates is between 3.690 and 11.359.

Inference about a population variance

• As automotive part must be machined to close tolerances to be acceptable to customers. Production

specifications call for a maximum variance in the lengths of the parts of 0.0004. Suppose the sample

2