BASIC PRACTICE OF STATISTICS+LAUNCHPAD
BASIC PRACTICE OF STATISTICS+LAUNCHPAD
8th Edition
ISBN: 9781319053093
Author: Moore
Publisher: MAC HIGHER
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Chapter 24, Problem 24.63SE
To determine

To find: The value of x and s for the insulin concentrations in the two types of mice.

To test: Whether there is a significant difference in the mean insulin concentrations in the two types of mice or not.

To check: The P-value confirm the claim in the report that P<0.005 .

Expert Solution & Answer
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Answer to Problem 24.63SE

The mean for Wild type is 59 and the mean for aP2/ is 0.75.

The value of s for Wild type is 2.85.

The value of s for aP2/ is 0.632.

The conclusion is that there is a significant difference in the mean insulin concentrations in the two types of mice.

Yes, the P-value confirm the claim in the report that P<0.005 .

Explanation of Solution

Given info:

The data shows that the insulin concentrations in the two types of mice.

In the given information, nWild=10 , naP2/=10 , x¯Wild=5.9 , x¯aP2/=0.75 , SEMWild=0.9 and SEMaP2/=0.2 .

Calculation:

Mean:

From the information in Exercise 62, the mean for Wild type is 59 and the mean for aP2/ is 0.75.

Standard deviation for Wild type:

SEMWild=sWildnWildsWild=SEMWildnWild=0.9(10)=2.85

Thus, the value of s for Wild type is 2.85.

Standard deviation for aP2/ :

SEMaP2/=saP2/naP2/saP2/=SEMaP2/naP2/=0.2(10)=0.632

Thus, the value of s for aP2/ is 0.632.

PLAN:

Check whether or not there is a significant difference in the mean insulin concentrations in the two types of mice.

State the test hypotheses.

Null hypothesis:

H0:μ1=μ2

Alternative hypothesis:

Ha:μ1μ2

SOLVE:

Test statistic and P-value:

Software procedure:

Step by step procedure to obtain test statistic and P-value using the MINITAB software:

  • Choose Stat > Basic Statistics > 2-Sample t.
  • Choose Summarized data.
  • In first, enter Sample size as 10, Mean as 5.9, Standard deviation as 2.85.
  • In second, enter Sample size as 10, Mean as 0.75, Standard deviation as 0.632.
  • Choose Options.
  • In Confidence level, enter 95.
  • In Alternative, select not equal.
  • Click OK in all the dialogue boxes.

Output using the MINITAB software is given below:

BASIC PRACTICE OF STATISTICS+LAUNCHPAD, Chapter 24, Problem 24.63SE

From the MINITAB output, the value of the t-statistic is 5.58 and the P-value is 0.000.

CONCLUDE:

The P-value is 0.000 and the significance level is 0.05.

Here, the P-value is less than the significance level.

That is, 0.000(=P-value)<0.05(=α) .

Therefore, by the rejection rule, it can be concluded that there is evidence to reject H0 at α=0.05 .

Thus, there is a significant difference in the mean insulin concentrations in the two types of mice.

Justification:

From the MINITAB output, the P-value is 0.000. Hence, the P-value confirm the claim in the report that P<0.005 .

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Chapter 24 Solutions

BASIC PRACTICE OF STATISTICS+LAUNCHPAD

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