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Confidence Interval of the Mean , Independent-, and Paired-Samples T-tests

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### Confidence Interval of the Mean , Independent-, and Paired-Samples T-tests

### Confidence Interval of the Mean

### Confidence Interval of the Mean

### Confidence Interval of the Mean

We often use limited observations (samples) to talk about or estimate the population values from which they come.

For example: My driver friend wants to know if he should take the F train; approximately how frequently does this train arrive?

If I tell him approximately 8 minutes, how good is this estimate? How justified am I in using my SAMPLE mean here?

Confidence Interval of the Mean (95% or 99%)

95 % Confidence Interval of the Mean

8

+ 2.57

(5.10/√6)

= 13.35

8

- 2.57

(5.10/√6)

= 2.65

There is a 95% probability that the TRUE population mean is between 2.65 and 13.35 minutes.

99% Confidence Interval of the Mean

8

+ 4.03

(5.10/√6)

= 16.39

8

- 4.03

(5.10/√6)

= -.39

There is a 99% probability that the TRUE population mean is between -.39 and 16.39 minutes.

Confidence Interval of the Mean

sample mean

Standard error

t critical value

(look up in table)

If 95% CI, use a=.05

If 99% CI, use a=.01

Remember, df = N-1

Seven people are recruited to test Proactiv acne treatment. Each person’s face is examined by a dermatologist who reports the number of pimples on each person’s face. Individuals are then instructed to use the Proactiv system of products for 3 months, after which they return to have their face pimples counted again. Test the hypothesis that Proactiv produces a difference in pimple number, using an alpha level of .05.

Step 1: State the null and alternative hypotheses:

H0: Proactiv does not produce a difference in pimple number.

H1: Proactiv produces a difference in pimple number.

Step 2: Find the critical value.

Two-tailed, alpha .05, df = 6

tcrit = -2.45 and +2.45

Seven people are recruited to test Proactiv acne treatment. Each person’s face is examined by a dermatologist who reports the number of pimples on each person’s face. Individuals are then instructed to use the Proactiv system of products for 3 months, after which they return to have their face pimples counted again. Test the hypothesis that Proactiv produces a difference in pimple number, using an alpha level of .05.

Step 3: Calculate the obtained statistic:

-.88

____

=

Seven people are recruited to test Proactiv acne treatment. Each person’s face is examined by a dermatologist who reports the number of pimples on each person’s face. Individuals are then instructed to use the Proactiv system of products for 3 months, after which they return to have their face pimples counted again. Test the hypothesis that Proactiv produces a difference in pimple number, using an alpha level of .05.

Step 3: Calculate the obtained statistic:

-.88

____

=

= - 2.15

.40

Step 4: Make a decision.

I

I

-2.45

2.45

Retain the null hypothesis.

Experimental design:

Two separate groups, each experiencing a different treatment.

A food writer would like to review the pricing of cocktails in big cities. She is looking specifically to compare the price of cocktails in Boston and New York to examine whether or not the average cocktail price is different. She goes to 7 bars in Boston and 7 bars in New York, recording the price of each bar’s Cosmopolitan. Below is the data. Test the hypothesis that Boston and New York charge significantly different prices for cocktails using an alpha level of .05.

Step 1: State the null and alternative hypotheses:

H0: Boston and NY do not charge different prices for cocktails.

H1: Boston and NY do charge different prices for cocktails.

Step 2: Find the critical value.

For an independent-groups t-test, we use df = N-2

Alpha = .05, 2-tailed, df = 12

tcrit = -2.18 and +2.18

A food writer would like to review the pricing of cocktails in big cities. She is looking specifically to compare the price of cocktails in Boston and New York to examine whether or not the average cocktail price is different. She goes to 7 bars in Boston and 7 bars in New York, recording the price of each bar’s Cosmopolitan. Below is the data. Test the hypothesis that Boston and New York charge significantly different prices for cocktails using an alpha level of .05.

Step 3: Calculate the obtained statistic

5.86 – 6.71

__________

=

5.86 6.71

4.86 17.43

A food writer would like to review the pricing of cocktails in big cities. She is looking specifically to compare the price of cocktails in Boston and New York to examine whether or not the average cocktail price is different. She goes to 7 bars in Boston and 7 bars in New York, recording the price of each bar’s Cosmopolitan. Below is the data. Test the hypothesis that Boston and New York charge significantly different prices for cocktails using an alpha level of .05.

Step 3: Calculate the obtained statistic

5.86 – 6.71

__________

=

= -1.16

.73

Step 4: Make a decision

5.86 6.71

4.86 17.43

I

I

-2.18

2.18

Retain the null hypothesis.

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