Pharmacie sans ordonnance livraison rapide 24h: acheter viagra en ligne en France.

## Lesson 3.5

**Lesson 3.5 **
**Testing Hypothesis about two means: Independent and **

Large Samples.
** **

In this lesson you will learn about testing hypothesis about two population

means (µ1 and µ2) and constructing confidence intervals for the difference

(µ1- µ2) between two population means. In doing this we will make the

following assumptions:

1.

The samples are independent. This means that the sample collected from one population is not related to the sample collected from the other population.
The two sample sizes are large, n1 > 30 and n2 > 30. This means that the critical value (cv) and the test statistic (ts) both are z-values.
A good example of testing a hypothesis involving two populations is the case of testing a drug for side effects. Two groups of people are used in testing a drug. The first group is called the “treatment group” who are actually treated with the medicine and the other group is called the control group who actually do not use the drug. Here the treatment group which is the first sample is not related to the control group which is the second sample.
1 and n1 to represent the sample mean, sample
standard deviation and the sample size respectively. Similarly for sample 2

We use µ1 for the mean of the first population and µ2 for the mean of the

second population. The claim is always regarding µ1 and µ2, and we use the

two samples to test the claim. µ1 > µ2, µ1 < µ2, µ1 = µ2, µ1 ≠ µ2 are typical

claims using two populations.

The z-cv is obtained using the

**invNorm **menu in TI-83

The z-ts is given by:

(

*x *−

*x *) − (µ − µ )

We use

**2-SampZtest** menu in TI-83 to obtain z-ts. This menu can be

accessed by:

[STAT],TESTS,[3].

**We will use s1 and s2, the standard deviations of the samples in place of **
σ

**1 and **σ

**2 in the calculator**.

**Example 1:** The Anderson Pharmaceutical Company wants to test prilosec, a new

medicine for acid reflex disease. In order to test the effect of prilosec on

systolic blood pressure, 50 people with acid reflex disease are given prilosec

and 100 others are not. The systolic blood pressure is measured for each

subject and the sample statistic is as follows:

The head of research at Anderson claims that prilosec does not affect blood

pressure. Use a 0.01 level of significance to test this claim.

**Solution: **

Prilosec has no effect on systolic blood pressure means that the mean blood

pressure of those in the treatment group (µ1) equal to the mean blood

pressure of those in the control group (µ2).

1.

This is a two-tailed test. (tail area = α/2 = 0.005)
Using area to the left = 0.005 in the

**invNorm** menu gives:

** z-cv = **±

**2.58 **
Access the

**2-SampZTest** by pressing [STAT],right arrow twice to

choose TESTS, [3]. Now type in each required data, choose two-tailed

test and down arrow to Calculate and [ENTER]. The calculator

display is shown below. This gives the

**z-ts = 1.33**
z-cv, z-ts and the critical regions are shown in fig. 2 below.
The z-ts does not fall in the critical region, so we fail to reject the Ho:
Claim is the same as the null hypothesis and we fail to reject the Ho. There is not enough evidence to reject the claim that prilosec does not affect systolic blood pressure.

**Confidence Interval for the difference between the two means: **

A confidence interval for µ1 -µ2 is of the form:

(

*x *−

*x *) −

*E *< µ − µ < (

*x *−

*x *) +

*E *
where E is the margin or error and is given by:
As the sample sizes are large, we replace σ1 and σ2 by s1 and s2 respectively.

**Example 2**:

Using the samples in example 1, construct the 99% confidence interval for
µ1 - µ2.

**Solution: **

1.

α = 0.01, therefore, α/2 = 0.005 and zα/2 = 2.58
(

*x *−

*x *) −

*E *< µ − µ < (

*x *−

*x *) +

*E*
8.0 −15.5 < µ − µ < 8.0 + 15.5
Since this interval contains zero, there is not enough evidence to reject the claim that µ1 – µ2 = 0

**Example 3:**
A study was conducted to compare the weights of college freshmen who commute from home and those who are resident on college campus. For 750 freshmen who commute from home, the mean is 72.5 kg and the standard deviation is 14.8 kg. The mean weight of 1250 resident freshmen is 68 kg with a standard deviation of 12.7 kg. (a)
Use a 0.01 level of significance to test the claim that the resident
freshmen come from a population with a mean that is less than the mean for the commuting freshmen. (b)
Construct the 99% confidence interval for the difference between the
means of the two population of students.

**Solution: **

(a):

The sample data can be summarized as follows:

The statement of claim is always made starting with population 1 1.
This is a right tail test. (tail area = α = 0.01)
Using area to the left = 1 – 0.01 in the

**invNorm** menu, we get:

**z-cv = 2.33 **
Using the sample data in the

**2-SampZTest** menu, we get:

**z-ts = 6.93**
**[STAT],TESTS,[3] **
The values of z-cv, z-ts and critical region are shown in the figure below.
Fig. 4. z-ts is inside the critical region
The z-ts is inside the critical region, and so we reject the null hypothesis.
The claim is the same as the alternative hypothesis, and we reject the null hypothesis. Therefore, we support the alternative hypothesis and hence the claim.

**There is enough evidence to support the claim that the resident **

freshmen come from a population with a mean that is less than

the mean for the commuting freshmen.

99% confidence gives α = 0.01, and α/2 = 0.005. Therefore zα/2 = 2.58
(

*x *−

*x *) −

*E *< µ − µ < (

*x *−

*x *) +

*E*
This interval supports the claim that µ1 > µ2.

**Example 4:**
A recent report in Charlotte Sun Herald claimed that Charlotte county students performed better in the math section of the 10th grade FCAT than students from Lee County. A sample of 35 10th graders from Charlotte has the following math scores.
79, 85, 92, 55, 62, 78, 89, 88, 59, 92, 85, 87, 91, 62, 69, 71, 51, 42, 69, 74, 53, 84, 91, 55, 69, 72, 82, 95, 60, 54, 42, 55, 65, 71, 80
A sample of 40 10th graders from Lee County has the following math scores.
88, 51, 42, 91, 75, 68, 44, 75, 78, 64, 62, 69, 92, 44, 38, 72, 69, 84, 89, 53, 49, 93, 99, 42, 53, 63, 74, 82, 51, 41, 32, 67, 75, 83, 79, 50, 44, 69, 70, 58
At the 0.04 significant level test the claim of the Sun Herald.
Construct the 96% confidence interval for the difference between the two mean scores.

**Solution: **
Clear existing lists L1 and L2.

Use the data to construct lists L1 and L2 and do

**1-Var Stats** on these to

obtain the sample data.

**[STAT],[ENTER] **
**[STAT],CALC,[ENTER],[2nd][1] **
**[STAT],CALC,[ENTER],[2nd][2] **
The data can be summarized as follows::
This is a right tailed test. (area to the left = 1 – 0.04)
Using the sample data in the

**2-SampZTest**, we get z-ts = 1.6

The values of z-cv, z-ts and the critical region are shown in the figure below.

Fig. 6. z-ts is not in the critical region.
The z-ts is not inside the critical region, so we fail to reject the null hypothesis.
The claim = H1: and we fail to reject the Ho:, therefore we fail to support H1 and hence the claim. There is not enough evidence to support the claim that Charlotte 10th graders performed better than Lee 10th graders.
(

*x *−

*x *) −

*E *< µ − µ < (

*x *−

*x *) +

*E*
6.11 − 7.82 < µ − µ < 6.11 + 7.82

**Inferences about Two Proportions. **
In this section we will test hypothesis made about two population proportions and construct confidence interval about the difference between the two population proportions. Here we will assume that the two samples are randomly selected and are independent. The following are the notations we will use in this section.
x1 = number of success in the sample x2 = the number of successes in the

Typical claims involving proportions are p1 > p2, p1 < p2, p1 = p2 and so on.

The critical value in tests involving proportion is the z-cv and it is obtained

using the

**invNorm **menu in TI-83.

The test statistic is the z-ts which is defined as follows ;

( ˆ

*p *− ˆ

*p *) − (

*p *−

*p *)

We will use the

**2-PropZTest** to obtain z-ts in this case. This menu can be

accessed by pressing [STAT],right arrow twice to select TESTS,[6]. Type in

the required data, choose the type of test and down arrow to choose calculate

and [ENTER].

**Confidence Intervals**:

The confidence interval for the difference between population proportions p1

– p2 is constructed as follows:

( ˆ

*p *− ˆ

*p *) −

*E *< (

*p *−

*p *) < ( ˆ

*p *− ˆ

*p *) +

*E * where E is given by:

**Example 5:**
A U.S. Department of Education report included the claim that “girls are less likely to be dropped out of school than boys before they are 18.” Sample data consisted of 1350 girls, 238 of them dropped out of school before they completed 18 and 2358 boys, 422 of them dropped out of school before they were 18.
Use a 0.05 significance level to test this claim.
Construct the 95% confidence interval for the difference between the two population proportions.

**Solution: **

Here population 1 consists of girls who dropped out of school before 18, and

population 2 consists of boys who dropped out of school before 18. The

claim is that the proportion of girls dropping out of school before 18 is less

than the proportion of boys dropping out of school before 18. The sample

data is as follows:

This is a left tail test. (Since no significance level is specified, we choose 0.05 and this is the tail area.)
Using area to the left = 0.05 in the

**invNorm** menu, we get:

**z-cv = -1.645 **
Use the sample data in the

**2-PropZTest** and obtain z-ts as shown in

the figure below.

**z-ts = - 0.204**
**[STAT],TESTS,[6] **
The values of z-cv, z-ts and the critical region are shown in the figure below.

**z-ts = - 0.204 **
**cv = - 1.645 **
Fig. 8. z-ts is outside the critical region.
The z-ts is outside the critical region, and so we fail to reject the null hypothesis.
Claim = H1 and we fail to reject Ho, and so we fail to support H1 and

hence the claim.

**There is not enough evidence to support the claim that proportion **

of girls dropping out of school is less than the proportion of boys

dropping out of school.
95% confidence level means that α = 0.05 and α/2 = 0.025.
Using area to the left as 0.025 in

**invNorm** menu, we get zα/2 = 1.96

ˆ

*p *− ˆ

*p *= 0.176 − 0.179 = −0.003
−

*E *< (

*p*1 −

*p *)
= - 0.003 – 0.026 < (p1 – p2) < - 0.003 + 0.026
This result supports the conclusion of the hypothesis test as p1 – p2 is likely

to contain zero.

** **

Example 6:

A public relations expert and consultant for the television industry is planning a strategy to influence voter perception of government regulation of television programs. In a survey it is found that 35% of 552 Democrats believe that the government should regulate television programs, compared to 41% of the 417 Republicans surveyed. (a)
At the 0.05 significance level, test the claim that there is no difference
between the proportions of Democrats and Republicans who believe in government regulation of television programs. (b)
Construct a 95% confidence interval for the difference between the
proportions of Democrats and Republicans who believe that government

should regulate television programs.

**Solution: **

There is no difference between the two proportions means that p1 = p2.

Summary of data:

This is a two-tailed test. Tail area = so α/2 = 0.025
Using area to the left = 0.025 in

**invNorm**, gives

**z-cv = **±

**1.96 **
Using the sample data in the

**2-PropZTest** menu, we get

**z-ts = -1.92** as shown in fig. 9 below.

**[STAT],TESTS,[6] **
The z-cv, z-ts and the critical region are shown in fig. 10 below.
Fig. 10. z-ts is not inside the critical region.
z-ts is not inside the critical region, so we fail to reject the null hypothesis.
Claim = Ho and we fail to reject Ho, so we fail to reject the claim.

**There is not enough evidence to reject the claim that the two **

proportions are the same.
95% confidence means that α = 0.05 and α/2 = 0.025
Using area to the left = 0.025 in

**invNorm** menu, we get zα/2 = 1.96

( ˆ

*p *− ˆ

*p *) −

*E *< (

*p *−

*p *) < ( ˆ

*p *− ˆ

*p *) +

*E *
= (0.35 – 0.41) - 0.062 < p1 – p2) < (0.35 – 0.41) + 0.062
=

**-0.122 < (p1 – p2) < 0.002**
Since the confidence interval contains zero, p1 – p2 = 0 is a strong

possibility.

**Example 7**:

A random sample of 216 registered voters in Florida showed that 125 of

them voted in the 2000 presidential elections. A random sample of 288

registered voters in Texas showed that 141 of them voted in the presidential

elections of 2000.

(a)

At a 5% significance level, test the claim that the proportion of registered voters who voted in Florida is greater than the proportion of registered voters who voted in Texas.
Construct the 95% confidence interval for the difference between the voter turnovers in Florida and in Texas.

**Solution: **

The data can be summarized as follows:

This is a right-tail test. Tail area = α = 0.05
Using area to the left = 1 – 0.05,

**z-cv = 1.64 **
Using the sample data in the

**2-PropZTest** menu,

**z-ts = 1.983**
**z-ts = 1.98 **
**z-cv = 1.64 **
z-ts is inside the critical region. Therefore, we reject the null hypothesis.
Claim = H1, and we reject the null hypothesis. Therefore, we support the alternate hypothesis and hence the claim. There is enough evidence to support the claim that the proportion of voter turnout in Florida is greater than the proportion of voter turnout in Texas.
Degree of confidence is 95% means that α = 0.05 and α/2 = 0.025
( ˆ

*p *− ˆ

*p *) −

*E *<

*p *−

*p *< ( ˆ

*p *− ˆ

*p *) +

*E*
(0.579 − 0.490) − 0.088 <

*p *−

*p *< (0.579 − 0.490) + 0.088
0.011 <

*p *−

*p *< 0.177

**Homework: **
Anderson Window manufacturing company suspects a difference in the mean number of sick leave taken by workers in the day shift compared to the night shift. A random sample of 35 day workers had an average sick leave of 11.8 days with a standard deviation of 3.8 days last year. A random sample of 45 night workers had an average sick leave of 16.5 days with a standard deviation of 4.2 days. (a)
At the 0.05 level of significance, test the company’s claim that
the night workers take more sick leave than day workers. (b)
Construct the 95% confidence interval for difference between
the mean sick leaves of the day and night workers. Does the confidence interval agree with the conclusion of the hypothesis test? Comment.
A random sample of 52 winter days in Los Angeles gave a mean pollution index of 48 with a standard deviation of 21. A random sample of 35winter days in New York gave a mean pollution index of 28 with a standard deviation of 12. (a)
At the 0.1 level of significance, test the claim that the mean
pollution index for Los Angeles is greater than the mean pollution index for New York. (b)
Construct an 90% confidence interval for the difference
between the average pollution indices for Los Angeles and New York.
Each night a person has both REM (rapid eye movement) and non-REM sleep. It is believed that children have more REM sleep than adults. Observation of a sample of 38 5-year old children showed that
they had an average REM sleep of 3.2 hours with a standard deviation of 1.3 hours. Observation of a sample of 42 adults showed that they had an average REM sleep of 2.9 hours with a standard deviation of 1.2 hours. (a)
At the 0.04 significant level, test the claim that children have
Construct the 96% confidence interval for the difference
between the mean REM sleep of children and adults.
A sales person tells you that the average repair cost for model A vacuum cleaners is the same as the average repair cost for model B vacuum cleaners. A sample of 42 model A vacuum cleaners have a mean repair cost of $52.00 and a standard deviation of $8.00. A sample of 38 model B vacuum cleaners has a mean repair cost of $49.00 with a standard deviation of $9.5. (a)
Use a 5% significance level to test the sales person’s claim.
Construct a 95% confidence interval for the difference between
A psychologist believes that among high school seniors, girls spend more time studying than boys. The number of hours per week studied by a random sample of 35 girls is given below. 18.5, 27.4, 16.1, 19.9, 20.9, 17.6, 15.9, 21.8, 23.0, 20.6, 20.5, 31.0, 24.0, 14.4, 22.5, 21.6, 12.0, 15.6, 22.8, 27.6, 22.1, 18.5, 25.5, 22.5, 20.0, 27.8, 13.2, 18.4, 21.6, 23.0, 24.5, 28.0, 17.9, 16.2, 24.8 The number of hours per week studied by a random sample of 32 boys is given below. 16.8, 15.9, 12.7, 21.8, 13.9, 12.9, 15.8, 11.9, 17.3, 12.1, 14.6, 13.5, 17.4, 19.8, 18.9, 15.7, 14.8, 19.1, 7.9, 10.6, 23.4, 15.8, 14.9, 19.8, 21.7, 17.2, 10.9, 7.8, 16.3, 21.6, 23.0, 24.3. (a)
At the 0.05 significance level, test the claim that girls study
more number of hours per week than boys. (b)
Construct the 95% confidence interval for the difference
A random sample of 1200 adults in Florida shows that 19% of them are smokers. A random sample of 1500 adults in Georgia shows that 22% are smokers. (a)
At the 0.05 level of significance, test the claim that the
proportion of adult smokers is the same in Florida and in Georgia.
Construct the 95% confidence interval for the difference
between the proportions of smokers in these two states.
A random sample of 135 adults with no college education showed that 28 of them believe in extraterrestrial life. A random sample of 220 college graduates showed that 42 of them believe in extraterrestrial life. At the 0.08 level of significance, test the claim that the proportion of people with no college degrees who believe in extraterrestrial life is different from the proportion of college graduates who believe in extraterrestrial life.
A survey of people’s willingness to donate organs produced the following results. A random sample of 130 men showed that 81 of them have signed up for an organ donor program. A random sample of 112 women showed that 56 of them have signed up for organ donation. (a)
At the 0.05 level of significance, test the claim that the
proportion of men organ donors is greater than the proportion of women organ donors (b)
Construct the 95% confidence interval for the difference
between the proportions of men organ donors and women organ donors.
Some parents believe that private school students generally score higher in the SAT than public school students. A survey of 1250 public school students who took SAT showed that 458 of them scored above 1000. Another survey of 885 private school students showed that 382 of them scored above 1000. At the 0.05 significance level, test the claim that the proportion of public school students who score above 1000 in SAT is less than the proportion of private school students who score above 1000.
In a survey of 2200 AT&T customers, 1850 of them said they are satisfied with their long distance telephone service. Another survey of 1860 MCI customers, 1237 said they are satisfied with their long distance service. At the 0.02 level of significance, test the claim that the proportion of satisfied customers is the same in both cases.

**PROJECT: **

You are required to undertake a comparative study about the proportion of high school graduates who go to college from charlotte High and Port charlotte High. (You can pick two other schools of your choice). (i)
Collect appropriate data. (The school guidance counselor will be able to provide you with the data that you need).
(iii) Organize and summarize your data. (iv) Do a 10-step hypothesis test. (v)
Make a clear statement about your conclusion.

Source: http://gmanacheril.com/STA2023/Text%20Materials/Lesson%203.5.pdf

In situ hybridisation on yeast with Locked Nucleic Acid in situ detection probesProtocol prepared by Dr. Torben Heick, Department of Molecular Biology Aarhus University, DenmarkThe protocol originates from the Singer laboratory (Long et al. (1997) Science 277, 383-387) and has been modified for yeast, Saccharomyces cerevisiae, cell cultures subjected to a heat-shock treatment. However, the outli

PE7734038-1 Naproxen Sodium Tablets 275-550mg Cut Size : 270 x 420 mm 6 pt helv condensed Pharma code: 640 Date: 11.03.2008 (Front) Naproxen Naproxen Sodium Tablets USP • For the relief of the signs and symptoms of rheumatoid arthritis Medication Guide • For the relief of the signs and symptoms of osteoarthritis • For the relief of the signs and symptoms of ankylosing spondyli