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Zea Mays. Refer to Exercise 10.123.

a. Determine a 95% confidence interval for the difference between the mean heights of cross-fertilized and self-fertilized Zea mays.

b. Repeat part (a) for a 99% confidence level.

Short Answer

Expert verified

a).(0.0303,41.8297)is the 95%confidence interval for the data

b). (-8.0776,49.9376) is the 99% confidence interval for the data.

Step by step solution

01

Part (a) Step 1: Given Information

The values are d¯=20.93,sd=37.74, the level of significance is 0.05, the data is shown below.

49
-67 816 6
23
28 4114
29
56 24
7560
-48
02

Part (a) Step 2: Explanation

The degree of freedom is,

dof=n-1

=15-1

=14

The level of significant crucial value is ±2.145.

Therefore, we can compute the confidence level as follow:

CI=d¯±ta2sdn

=20.93±(2.145)37.7415

=(0.0303,41.8297)

03

Part (b) Step 1: Given Information

The values are d¯=20.93,sd=37.74,

the significance level is 0.05, the data is shown as follow.

49
-67
8
16
6
23
28
41
14
29
56
24
75
60
-48
04

Part (b) Step 2: Explanation

Let's compute the degree of freedom:

dof=n-1

=15-1

=14

The level of significant crucial value is ±2.977.

Therefore, we can compute the confidence level as follow,

CI=d¯±ta2sdn

=20.93±(2.977)37.7415

=(-8.0776,49.9376)

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