dependent and independent samples

Dependent and Independent Samples

1Section 10.1, Page 208

Problems

2Problems, 30

Problems

3Problems, Page 230

Confidence Interval – Dependent SamplesIllustrative Problem

sample mean = 6.33 , Sx = 5.13, n = 6

Find the 95% confidence interval for the mean of the difference? (Sampling Distribution is t-distribution, 5df)C.I. = sample mean ± Margin of Error =sample mean ± critical value * standard error6.33 ± 2.5706 * 6.33 ± 5.38 = (.95, 11.71)

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5.13/ 6


Confidence Interval – Dependent SamplesIllustrative Problem TI-83 Black Box Program

Find the 95% Confidence Interval d = B – A(We have to make the assumption that the data is taken from a normal population.) STAT – ENTER: Enter Brand A in L1 and B is L2Highlight the title box for L3.=2nd L2 – 2nd L1 Enter (Difference set is now L3).STAT–TESTS- 8:TIntervalComplete procedure as described in Section 9.1

Confidence Interval is (.96, 11.71)


Hypothesis Test– Dependent SamplesIllustrative Problem

sample mean = 6.33 , Sx = 5.13, n = 6

Test the hypotheses that the tread wear for Brand B is greater than for Brand A.

Ho: μd = B – A =0 (No difference)Ha: μd = B – A > 0 (B is greater)

μd=0

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d = 6.33

p-value = .0146

p-value = PRGM – TDISTLOWER BOUND = 6.33UPPER BOUND = 2ND EE99MEAN = 0STANDARD ERROR = df = 5Answer: p-value = .0147 (Reject Ho, B has greater tread wear.)

6Section 10.2, Page 211€

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Hypothesis Test– Dependent SamplesIllustrative Problem – TI-83 Black Box Program

Test the hypotheses that the tread wear for Brand B is greater than for Brand A. (d=Brand B – Brand A)

Ho: μd = 0 (No difference in wear)Ha: μd > 0 (Brand B has more wear)

STAT – ENTER: Enter Brand A in L1 and B in L2Highlight L3=2nd L2 – 2nd L3 Enter (L3 in now difference set)STAT-TESTS– 2:T-TestComplete the program as described in Section 9.2 with L3 as the difference set.

P-Value = .0146 (Reject H0, Brand B has greater tread wear.)


Problems

a. Test the hypotheses that the people increased their knowledge. Use α=.05 and assume normality. State the appropriate hypotheses.

b. Find the p-value and state your conclusion.

c. Find the 90% confidence interval for the mean estimate of the increase in test scores.

8Problems, Page 231

Problems

a. State the appropriate hypotheses.b. Find the p-value and state your

conclusion.c. Find the 98% confidence interval for

the mean estimate of the increase in number of sit-ups.

9Problems, Page 232

Independent SamplesTwo Means: Unknown σ

Suppose we want to compare the income of Shoreline students with UW students.

From sample of 50 Shoreline Students:

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x s = $12,500

From sample of 60 UW students:

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x UW = $14,750

Test the hypothesis that the populations incomes are not the same.

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H0 : μUW = μS ⇒ μUW − μS = 0

Ha : μUW ≠ μS ⇒ μUW − μS ≠ 0

To find a p-value we need a sampling distribution for

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x UW − x S


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SS = $4,300

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SUW = $4,800


μ1 σ1 μ2 σ2

Population 1 Population 2

Sampling Dist.

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x 1 Sampling Dist.

€

x 2

Sampling Dist.

€

x 1 − x 2

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ux 1 −x 2= μ1 − μ2

σ x 1 −x 2= (σ x 1

)2 + (σ x 2)2 =

S12

n1

+S2

2

n2

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μx 1= μ1

σ x 1=

S1

n1

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μx 2= μ2

σ x 2=

S2

n2


A T-Distribution Applies


Conditions for t-distribution model for sampling distribution:

1.The samples are independent of each other – two unrelated sets of sources are used, one from each population.

2.The populations are normal, or the sample size from each population is large, n ≥ 30

The correct degrees of freedom for the sampling distribution for the difference of two independent means is given by the following formula:

The calculator will make this calculation.


Independent Means – Illustrative ProblemConfidence Interval – TI-83 Add-in

Find the 95% confidence interval for the difference in the heights, μm – μf.

C. I. = sample mean difference ± Margin of Error=sample mean difference ± critical value * standard errorPRGM – STDERROR – 5: 2 MEANSSx1 = 1.92; n1=30; Sx2 = 2.18; n2 = 20Answer: SE = .6004; df = 37.2121PRGM – CRITVAL – 2C – LEVEL = .95; df (INTEGER) = 37Answer: 2.0262

C.I. = 69.8-63.8 ± 2.0262 * .6004 = (4.78, 7.22)


Independent Means – Illustrative ProblemConfidence Interval – TI-83 Black Box

Find the 95% confidence interval for the difference in the heights, μm – μf.

STAT – TESTS – 0:2 – SampTIntInpt: Stats : 69.8; Sx1: 1.92; n1: 30 : 63.8; Sx2: 2.18; n2: 20C-Level: .95Pooled: NoCalculateAnswer: (4.78, 7.22)


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x 1

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x 2

Independent Means – Illustrative ProblemHypotheses Test– Add-In Programs

Is this sufficient evidence to prove that male students are taller than female students.

Ho: μm – μf = 0 μm = μfHa: μm – μf > 0 μm > μf


PRGMS – TDISTLOWER BOUND = 6UPPER BOUND = 2ND EE99MEAN = 0STANDARD ERROR = .6004 (Slide 12)df = 37.2121 (Slide 13)Answer: p-value = 2.1930E-12 ≅ 0(Reject Ho, there is sufficient evidence to prove males are taller.)

0

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x m − x f = 6

p-value ≅ 0

Independent Means – Illustrative ProblemHypotheses Test– TI-83 Black Box

Is this sufficient evidence to prove that male students are taller than female students.

Ho: μm – μf = 0 μm = μfHa: μm – μf > 0 μm > μf

STAT – TESTS – 4: 2 – SampTTestInput: Stats :69.8; Sx1:1.92; n1:30 :63.8; Sx2:2.18; n2:20μ1: >μ2Pooled: NoCalculate; Answer: p-value = 2.1947E-12≅0(Reject Ho, there is sufficient evidence to prove males are taller)

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x 1

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x 2


Problems

a. Write the necessary hypotheses.b. State the p-value and your decision.c. If you make an error, what type is it?d. Construct a 98% confidence interval for the

difference mean selling prices (North of Cedar – Provo)

17Problems, Page 233

Problems

a. Write the necessary hypotheses.b. State the p-value and your decision.c. If you make an error, what type is it?d. Construct a 98% confidence interval for the

difference mean weight gained (Diet B – Diet A)


Problems

Assume normality.

a. Is there convincing evidence that nonorgan donors have higher anxiety about death than organ donors? Write the hypotheses and find the p-value. State your conclusion.

b. Find the mean and standard error of the sampling distribution.

c. Find the 98% confidence interval for the means; nondonor – donor.


Problems

a. State the hypothesis (Assume Normality)b. Find the p-value, and state you conclusion.c. Find the mean and standard error of the sampling

distributiond. Find the 95% confidence interval for the

difference of the means; Gouda-Brie.


Independent SamplesTwo Proportions

Suppose we want to compare the proportion of students getting financial aid at Shoreline and UW.

From sample of 50 Shoreline Students:

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pS' = 0.61

From sample of 60 UW students:

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p'UW = 0.54

Test the hypothesis that the population proportions are not the same.

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H0 : pUW = pS ⇒ pUW − pS = 0

Ha : pUW ≠ pS ⇒ pUW − pS ≠ 0

To find a p-value we need a sampling distribution for

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p'UW −p'S


Independent SamplesTwo Proportions: Unknown σp

p1 p2

Population 1 Population 2

Sampling Dist. p’1 Sampling Dist. p’2

Sampling Dist. p’1 – p’2

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up'1 − p'2= p1 − p2

σ p'1− p'2= (σ p'1

)2 + (σ p '2)2 =

p'1 (1− p'1 )

n1

+p'2 (1− p'2 )

n2

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μp'1= p1

σ p'1=

p'1 (1− p'1 )

n1

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μp'2= p2

σ p'2=

p'2 (1− p'2 )

n2

22Section 10.4, 1

The Normal Model Applies

Two ProportionsConditions for Normal Model

The sampling distribution for the difference of two sample proportions is a normal model if:

Samples are independent of each other

Each sample size > 20

Each sample has more than 5 successes (np or np’) and 5 failures (nq or nq’). (q=1-p)

Each sample is not more than 10% of its population


Two Proportions – Confidence Interval Illustrative Problem - TI-83 Add-In

A campaign manager wants to estimate the difference in support between women and men for his candidate. A sample of 1000 for each population was taken, and 459 women and 388 men favored his candidate. Find the 99% confidence interval for pw – pm.

C.I. = sample proportion difference ± Margin of Error =sample proportion difference ± critical value * standard error

PRGM - CRITVAL – 1:NORMAL DIST – CONF LEVEL = .99Answer: 2.5758

PRGM – STDERROR- 2: 2 PROP: p1 = 459/1000; n1 = 1000;p2 = 388/1000; n2 = 1000Answer: sample proportion difference = .0710; standard error = .0220

C.I. = .0710 ± 2.5758*.0220 = (.0143, .1277)

We are 99% confident that the proportion of female support exceeds the true proportion of male support is between 1.43% and 12.77%.


Two Proportions – Confidence Interval Illustrative Problem - TI-83 TI-83 Black Box

A campaign manager wants to estimate the difference in support between men and women for his candidate. A sample of 1000 for each population was taken, and 459 women and 388 men favored his candidate. Find the 99% confidence interval for pw – pm.

STAT – TESTS – B:2PropZIntx1 = 459n1= 1000x2 = 388n2 = 1000C-Level = .99CalculateAnswer: (.0142, .1278)


Section 10.4, Page 223 26

Difference between 2 Population Proportions

For a hypothesis test, the null hypothesis is: Ho: p1 = p2. If the two proportions are equal, then the standard deviation of the proportions must be also equal. Since p’1 seldom equals p’2, the σ(p’1-p’2) does not

reflect this fact.

To get around this problem for hypothesis tests, we calculate the weighted average of p’1 and p’2, called p’p, or pooled sample proportion, and substitute is value for both p’1 and p’2 in the standard error calculation.

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p'p =Combined Successes

Combined Sample size=

n1p'1 +n2 p'2n1 + n2

σ p' p=

p'p (1− p'p )

n1

+p'p (1− p'p )

n2

Two Proportions – Hypotheses TestIllustrative Problem - TI-83 TI-83 Add-In

A telephone salesman claims that his phones are better than the competition, and that no more of his phones are defective than the competition. A sample is taken. Can we reject the salesman’s claim? Use α = .05.

Ho: ps = pc

Ha: ps > pc

PRGM – STDERROR – 3: 2-PROP POOLEDp1 = 15/150; n1 = 150; p2 = 6/150; n2 = 150Answer: p’s – p’c = .06; Standard Error = .0295

0 .06

p-value = .0210

PRGM – NORMDIST – 1LB = .06; UB = 2nd EE99; MEAN = 0; SE( )= .0295Answer: p-value = .0210. (Reject Null Hypotheses, and reject the salesman’s claim.)


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Two Proportions – Hypotheses TestIllustrative Problem - TI-83 TI-83 Black Box

A telephone salesman claims that his phones are better than the competition, and that no more of his phones are defective than the competition. A sample is taken. Can we reject the salesman’s claim? Use α = .05.

Ho: ps = pc

Ha: ps > pc

STAT – TESTS – 6:2PropZTestx1: 15n1: 150x2: 6n2: 150p1: >p2CalculateAnswer: p-value = .0208 (Reject the Null Hypotheses and the salesman’s claim.)


Problems

a. Test the hypotheses that the preference in city A is greater than the preference in city B. State the appropriate hypotheses.

b. Find the p-value and state your conclusion.c. What model is used for the sampling

distribution and what is the mean of the sampling distribution and its standard error?

d. Find the 97% confidence interval for the difference in preferences, city A – city B.


Problems

a. State the appropriate hypotheses.b. Find the p-value and state your conclusion.c. What model is used for the sampling distribution and

what is the mean of the sampling distribution and its standard error?

d. Find the 98% confidence interval for the difference in proportions, men – women.


dependent and independent samples

Documents

brand b brand aho

sampling distribution

critical value

mean estimate

greater tread wear

independent samplestwo

l2highlight l3

tdistribution model