test hypothesis

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Hypothesis Tests Developing Null and Alternative Hypotheses Type I and Type II Errors

Hypothesis Tests for Population Mean: s Known Hypothesis Tests for Population Mean: s Unknown

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One-tailed(lower-tail)

One-tailed(upper-tail)

Two-tailed

0 0: H

0: aH 0 0: H

0: aH 0 0: H

0: aH

Summary of Forms for Null and Alternative Hypotheses about a

Population Mean The equality part of the hypotheses always appears in the null hypothesis. In general, a hypothesis test about the value of a population mean must take one of the following three forms (where 0 is the hypothesized value of the population mean).

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Type I and Type II Errors

CorrectDecision Type II Error

CorrectDecisionType I Error

Reject H0

(Conclude > µ0)

Accept H0

(Conclude < µ0)

H0 True( < µ0)

H0 False( > µ0)Conclusion

Population Condition

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Two Basic Approaches to Hypothesis Testing

There are two basic approaches to conducting a hypothesis test:

1- p-Value Approach, and 2- Critical Value Approach

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1- p-Value Approach toOne-Tailed Hypothesis Testing

Reject H0 if the p-value <

In order to accept or reject the null hypothesis the p-value is computed using the test statistic --Actual Z value.

Do not reject (accept) H0 if the p-value >

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2- Critical Value Approach One-Tailed Hypothesis Testing

Use the Z table to find the critical Z value, and

Use the equation to find the actual Z--Z . The rejection rule is:• Lower tail: Reject H0 if Actual z < Critical -z• Upper tail: Reject H0 if Actual z > Critical z

In other words, if the actual Z (Z ) is in the rejection region, then reject the null hypothesis.

s

/

xzn

Equation for finding the actual Z

value:

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Steps of Hypothesis Testing

Step 1. Develop the null and alternative hypotheses.Step 2. Specify and n.

Step 3. Compute critical Z and actual Z values.

Step 4. Use either of the following approaches to make conclusion:1- p-Value Approach, or

2- Critical Approach

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Example: Metro EMS

The EMS director wants toperform a hypothesis test, with a0.05 level of significance, to determinewhether the service goal of the response time to be at most 12 minutes or less is being achieved.

The response times for a randomsample of 40 medical emergencieswere tabulated. The sample meanis 13.25 minutes. The populationstandard deviation is believed tobe 3.2 minutes.

One-Tailed Tests About a Population Mean:s Known

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1. Develop the hypotheses.

2. Level of significance and sample size are:= .05n = 40

H0: Ha:

p -Value and Critical Value Approaches

One-Tailed Tests About a Population Mean:s Known: Solution

3. Compute the value of the test statistic.

s

13.25 12 2.47/ 3.2/ 40

xzn

Actual z

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5. Make conclusion about H0

We are at least 95% confident that Metro EMS is not meeting the response

goal of 12 minutes.

p –Value Approach

4. Compute the p –value.From the Z table the actual z = 2.47

p–value = 0.5 - .4932 = .0068

Because p–value = .0068 < = .05, we reject H0.

One-Tailed Tests About a Population Mean:s Known: Solution Continued

using Z table,

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p-value

0 Zc =1.645

= .05

z Za =2.47

Solution ContinuedBecause p–value = .0068 < = .05, we reject H0.

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5. Make conclusion about H0

We are at least 95% confident that Metro EMS is not meeting the response

goal of 12 minutes.

Because actual z = 2.47 > Critical z = 1.645

we reject H0.

Critical Value Approach

For = .05, z.05 = 1.6454. Determine the critical value and rejection rule.

Reject H0 if actual z > 1.645

Finding critical z value 0.5 – 0.05 = 0.45Then, from table

1.64 + 1.65 3.29 / 2 = 1.645

One-Tailed Tests About a Population Mean:s Known: Solution Continued

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Excel: SWStat

One-Tailed Tests About a Population Mean:s Known

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Excel: SWStat

One-Tailed Tests About a Population Mean:s Known

P Approach

Critical Approach

Because actual z = 2.47 > Critical z = 1.645 we reject H0, or

Because p–value = .0068 < α = .05, we reject H0

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Example: Glow Toothpaste

Two-Tailed Test for Population Mean: s Known

Quality assurance procedures call forthe continuation of the filling process if thesample results are consistent with the assumption thatthe mean filling weight for the population of toothpastetubes is 6 oz.; otherwise the process will be adjusted.

The production line for Glow toothpasteis designed to fill tubes with a mean weightof 6 oz. Periodically, a sample of 30 tubeswill be selected in order to check thefilling process.

oz.G

low

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Example Continued: Glow Toothpaste

Perform a hypothesis test, at the 0.03level of significance, to help determinewhether the filling process should continueoperating or be stopped and corrected.

Assume that a sample of 30 toothpastetubes provides a sample mean of 6.1 oz.The population standard deviation is believed to be 0.2 oz.

Two-Tailed Test for Population Mean: s Known

oz.G

low

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1. Determine the hypotheses.

2. Alpha and sample size are given

3. Compute the value of the test statistic.

= .03 and n=30

p –Value and Critical Value Approaches

H0: Ha: 6

Two-Tailed Tests About a Population Mean:s Known: Solution

s

0 6.1 6 2.74/ .2/ 30

xzn

Actual z

Glow

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5. Determine whether to reject or to accept H0.

p –Value Approach

4. Compute the p –value.

For actual z = 2.74, the probability = 0.4969, thus p–value = 2(0.5 – 0.4969) = 2 (0.0031) = 0.0062

Because p–value = .0062 < = .03, we reject H0. We are at least 97% confident that the

mean filling weight of the toothpaste tubes is not 6 oz.

Two-Tailed Tests About a Population Mean:s Known: Solution Continued

Glow

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Glow

/2 = .015

0z/2 = 2.17

z

/2 = .015

-z/2 = -2.17za = 2.74z = -2.74

1/2p -value= .0031

1/2p -value= .0031

Solution ContinuedBecause p–value = .0062 < = .03, we reject H0.

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Two-Tailed Tests About a Population Mean:s Known: Solution Continued

Critical Value Approach To Find the Critical Z Value:

/2 = .015

Given that = 0.03, thus /2 = .015 and 0.5 – 0.015 = 0.485

Then from the table we need to find the z value of 0.485.

Locate 0.485 in the Z Table. Thus, the critical z value for 0.485

is 2.17

Critical z/2 = 2.17

Critical z/2

0.485

0.5

Glow

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Critical Value Approach

Conclusion:

We are at least 97% confident that the mean filling weight of the toothpaste tubes is not 6 oz.

Because actual z of 2.74 > critical z of 2.17, we reject H0

Two-Tailed Tests About a Population Mean:s Known: Solution Continued

Glow

s

0 6.1 6 2.74/ .2/ 30

xzn

Actual Z

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/2 = .015

0 2.17

Reject H0

z

Reject H0

-2.17

Actual Z Value

z xn

s0

/

/2 = .015

Critical Approach: Solution Continued

= 2.74

Critical Z values

Because actual z of 2.74 > critical z of 2.17, we reject H0.

Glow

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Excel: SWStatGlo

w

Two-Tailed Tests About a Population Mean:s Known

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Excel: SWStat

Two-Tailed Tests About a Population Mean:s Known

P Approach

Critical Approach

Glow

THOUGHT

Confidence Intervals Versus Hypothesis Tests

A standard confidence interval is equivalent to a two-tail hypothesis test.

All two tails tests can be handled either as hypothesis tests or as confidence intervals.

The confidence interval has the appeal of providing a graphic feeling for how close the hypothesized value lies to the ends of confidence interval.

Rejection Rule: If the confidence intervaldoes not contain H0 , we reject H0.

32 males between the ages of 40 and 69 years with moderate carotid disease were tested at the Henry Hospital over 39-,months period. Their mean systolic pressure was 146.6 mmHg with a standard deviation of 17.3 mmHg. At a = 0.05, is this sample consistent with a population mean of 140 mmHg, which is considered a borderline for dangerously high blood pressure (note: recent medical evidence suggests 130 as a borderline, but we will use the older benchmark)?

Thinking Challenge Example

Apply confidence interval approach to test the hypothesis

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Confidence Interval Approach: For this problem, the two-sided hypothesis would

be:H0: 4Ha:4

The 95% confidence interval (α=0.05) for is: x t s

n /2

Since interval 140.36 < < 152.84 does not contain =140 we would reject the hypothesis H0: 4 in favor of Ha:4

Margin of Error

Thinking Challenge ExampleSolution

146 – + (2.040) 17.3 /5.657

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Test Statistic

Hypothesis Tests About a Population Mean:

s Unknown

t xs n

0

/

This test statistic has a t distribution with n - 1 degrees of freedom.

Actual t Value

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A State Highway Patrol periodically samplesvehicle speeds at various locationson a particular roadway. The sample of vehicle speedsis used to test the hypothesis

Example: Highway Patrol One-Tailed Test About a Population Mean: s Unknown

The locations where H0 is rejected are deemedthe best locations for radar traps.

H0: < 65

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Example Continued: Highway Patrol

At Location F on I-75, a sample of 64 vehicles shows amean speed of 66.2 mph with asample standard deviation of4.2 mph. Use = .05 totest the hypothesis.

Use Excel

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Using SWStat

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Solution Using SWStat

P Approach

Critical Approach

Since p=0.0128 < α=0.05 we reject

H0 The locations where H0

is rejected are deemedthe best locations for radar traps.

H0: < 65

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0 critical t = 1.669

Reject H0

Do Not Reject H0

t

One-Tailed Test About a Population Mean:s Unknown: Solution Continued

t Statistic = Actual t = 2.286

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The current rate for producing 5 amp fuses at Ariana Electric Co. is 250 per hour. A new machine has been purchased and installed that, according to the supplier, will increase the production rate. The production hours are normally distributed. A sample of 10 randomly selected hours from last month revealed that the mean hourly production on the new machine was 256 units, with a sample standard deviation of 6 per hour.

At the .05 significance level can Ariana Electric Co. conclude that the new

machine is faster?

Thinking Challenge and

Solution

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Step 4 State the decision rule.There are 10 – 1 = 9 degrees of freedom.

Step 1 State the null and

alternate hypotheses.H0: µ < 250H1: µ > 250

Step 2 Select the level of

significance. It is .05.

Step 3 Find a test statistic. Use the t distribution since s is not known and n < 30.

The null hypothesis is rejected if t > 1.833 or, using the p-value, the null hypothesis is rejected if p ≤ 0.05

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162.3106250256

ns

Xt

Computed t (or actual t) of 3.162 > critical t of 1.833 and

From Excel, p of .0058 < So we reject Ho

The p(t >3.162) is .0058 for a one-tailed test.

Step 5 Make a decision and interpret the

results.

ConclusionThe mean number of

amps produced by the new machine is more

than 250 per hour.

Actual t

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Solution Using SWStat

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Solution Continued

Since computed t (or actual t)

of 3.162 > critical t of 1.833 and since p of .0058 < thus, we reject HoHence, we conclude that the

mean number of amps

produced by the new machine

is more than 250 per hour.

H0: µ < 250

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A group of young businesswomen wish to open a high fashion boutique in a vacant store, but only if the average income of households in the area is more than $45,000. A random sample of 9 households showed the following results.

Thinking Challenge

$48,000 $44,000 $46,000$43,000 $47,000 $46,000$44,000 $42,000 $45,000

andSolution

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Use the statistical techniques in Excel (SWStat) to advise the group on whether or not they should locate the boutique in this store. Use a 0.05 level of significance. (Assume the population is normally distributed).

Thinking Challenge (Continued)

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Thinking Challenge 4 (Solution)

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Summary of Selecting an Appropriate Test Statistic for

a Test about a Population Mean

End of Chapter 10

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