Question

In: Statistics and Probability

Referring to this data: 2.58     2.51     4.04     6.43     1.58     4.32    ...

Referring to this data:

2.58     2.51     4.04     6.43     1.58     4.32     2.2       4.19    

4.79     6.2       1.52     1.38     3.87     4.54     5.12     5.15

5.5       5.92     4.56     2.46     6.9       1.47     2.11     2.32

6.75 5.84     8.8       7.4       4.72     3.62     2.46     8.75

Suppose we can only pass the quality test if there is evidence to suggest that the true mean ignition time is more than 0.04 seconds (or 4 hundredths of a second). please show work.

a. What is the null hypothesis for this test?

b. What is the alternative?

c. Conduct this hypothesis test at α=0.05 using the normal distribution. State all parts, including the critical value, test statistic, and conclusion in context of the problem.

d. Conduct this hypothesis test at α=0.05 using the t-distribution. State all parts, including the critical value, test statistic, and conclusion in context of the problem. Does your conclusion change?

Solutions

Expert Solution

SOLUTION-

WE USE MINITAB-16 FOR THE COMPUTATION PURPOSE

TO COMPUTE STANDARD DEVIATION FOR THE SAMPLE-

ENTER THE SAMPLE> STAT> BASIC STATISTICS> DISPLAY BASIC STATISTICS> UNDER 'STATISTICS', SELECT 'STANDARD DEVIATION'> OK

THE STANDARD DEVIATION = 2.104

1.) THE NULL HYPOTHESIS IS ,

2.) THE ALTERNATE HYPOTHESIS IS ,

3.) PERFORMING A ONE SAMPLE-Z TEST-

STAT> BASIC STATISTICS> ONE SAMPLE-Z> SELECT THE SAMPLE> SET STANDARD DEVIATION IS 2.104> SET THE HYPOTHESIZED MEAN AS 0.04> UNDER 'OPTIONS', SET THE CONFIDENCE LEVEL AS 95.0 AND ALTERNATE AS 'GREATER THAN'> OK

THE TEST STATISTIC IS 11.66 AND THE CORRESPONDING P-VALUE IS 0.000

AS P-VALUE<0.05, WE REJECT THE NULL HYPOTHESIS AND CONCLUDE THAT MEAN TIME IS MORE THAN 0.04

4.) PERFORMING A ONE SAMPLE-T TEST-

STAT> BASIC STATISTICS>ONE SAMPLE-T> SELECT THE SAMPLE> SET THE HYPOTHESIZED MEAN AS 0.04> UNDER 'OPTIONS', SET THE CONFIDENCE LEVEL AS 95.0 AND ALTERNATE AS 'GREATER THAN'> OK

THE TEST STATISTIC IS 11.66 AND THE CORRESPONDING P-VALUE IS 0.000

AS P-VALUE<0.05, WE REJECT THE NULL HYPOTHESIS AND CONCLUDE THAT MEAN TIME IS MORE THAN 0.04

****IN CASE OF DOUBT, COMMENT BELOW. ALSO LIKE THE SOLUTION, IF POSSIBLE.


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