Example Bowker and Lieberman (1972) gave the results for ...



Hypothesis Test for Comparing Two Population Means

Using Independent Random Samples (Pooled-t Test)

Example Bowker and Lieberman (1972) gave the results for tests of two thermostats used in irons that were made by an old supplier and a new supplier. Twenty three thermostats from each supplier were tested at a setting of 550 F. The actual temperature, measured with a thermocouple, are given below. Do the data provide evidence of differences in the mean temperatures for the thermostats from the two suppliers? Conduct test at α=.05.

new old

530.3 559.7

549.4 554.8

551.7 544.6

549.9 550.7

536.7 551.1

538.8 538.8

559.1 554.5

538.6 538.4

565.4 552.9

550.0 555.0

554.7 558.4

569.1 560.3

559.3 534.7

544.0 545.0

566.3 538.0

556.9 563.1

558.8 553.8

543.3 564.6

555.0 553.0

551.1 548.3

554.9 535.1

554.9 544.8

536.1 548.7

Entering the data in Minitab:

|New |Old |

|530.3 |559.7 |

|549.4 |554.8 |

|. |. |

|. |. |

|536.1 |548.7 |

Is Normality Assumption Reasonable?

In Minitab: STAT> BASIC STATISTICS> NORMALITY TEST> VARIABLE New

[pic]

Since the plotted points fall approximately along a straight line, we can conclude that the normal distribution assumption is reasonable.

In Minitab: STAT> BASIC STATISTICS> NORMALITY TEST> VARIABLE Old

[pic]

Since the plotted points fall approximately along a straight line, we can conclude that the normal distribution assumption is reasonable.

Is Equal Variances Assumption Reasonable?

IN MINITAB:

STAT> BASIC STATISTICS > 2 VARIANCES > EACH SAMPLE IS IN ITS OWN COLUMN >

Ho: [pic]= [pic]

Ha: [pic][pic][pic]

α=.05

Test

Method DF1 DF2 Statistic P-Value

Bonett 1 — 0.75 0.385

Levene 1 44 0.36 0.549

p-value > α thus we fail to reject H0 at α=.05. We can conclude that the assumption of equal variances is reasonable. So we can go ahead and conduct pooled-t test.

Ho: μnew=μold

Ha: μnew[pic]μold

MINITAB COMMANDS: STAT > BASIC STATISTICS > 2-SAMPLE t >

EACH SAMPLE IS IN ITS OWN COLUMN Sample 1 New Sample 2 Old > OPTIONS > Click ASSUME EQUAL VARIANCES > ALTERNATIVE > NOT EQUAL

Two-sample T for New vs Old

N Mean StDev SE Mean

New 23 551.1 10.3 2.1

Old 23 549.93 8.80 1.8

Difference = μ (New) - μ (Old)

Estimate for difference: 1.13

95% CI for difference: (-4.56, 6.82)

T-Test of difference = 0 (vs ≠): T-Value = 0.40 P-Value = 0.691 DF = 44

Both use Pooled StDev = 9.5745

p-value > α thus we fail to reject Ho at α=.05. We do not have sufficient evidence to conclude that there exists a difference in the mean temperatures for the thermostats from the two suppliers.

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