An experiment was concerned with the possible effect of a small lesion in a particular area mster's brain on the activity level of the animal. Thirty hamsters were selected at random and 28. of randomly divided into two groups give died i of 15 animals each. Group I was used as a control group n no lesion, whereas Group II was given a lesion in the designated area. Because three animals n Group II, the actual number turned out to be 12 for that group. After full recovery by the ning hamsters, e and fixed unit of time was recorded. The unit time. Did the lesion significantly reduce the activity of the hamsters (o 05)? ach was given access to a running wheel, and the number of revolutions per following data show activity in hundreds of revolutions per Group I Group II yı = 9.26 5.14 S 1.45 S2-2.81

Holt Mcdougal Larson Pre-algebra: Student Edition 2012
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ISBN:9780547587776
Author:HOLT MCDOUGAL
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Chapter11: Data Analysis And Probability
Section11.5: Interpreting Data
Problem 1C
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An experiment was concerned with the possible effect of a small lesion in a particular area
mster's brain on the activity level of the animal. Thirty hamsters were selected at random and
28.
of
randomly divided into two groups
give
died i
of 15 animals each. Group I was used as a control group
n no lesion, whereas Group II was given a lesion in the designated area. Because three animals
n Group II, the actual number turned out to be 12 for that group. After full recovery by the
ning hamsters, e
and
fixed unit of time was recorded. The
unit time. Did the lesion significantly reduce the activity of the hamsters (o 05)?
ach was given access to a running wheel, and the number of revolutions per
following data show activity in hundreds of revolutions per
Group I
Group II
yı = 9.26 5.14
S 1.45 S2-2.81
Transcribed Image Text:An experiment was concerned with the possible effect of a small lesion in a particular area mster's brain on the activity level of the animal. Thirty hamsters were selected at random and 28. of randomly divided into two groups give died i of 15 animals each. Group I was used as a control group n no lesion, whereas Group II was given a lesion in the designated area. Because three animals n Group II, the actual number turned out to be 12 for that group. After full recovery by the ning hamsters, e and fixed unit of time was recorded. The unit time. Did the lesion significantly reduce the activity of the hamsters (o 05)? ach was given access to a running wheel, and the number of revolutions per following data show activity in hundreds of revolutions per Group I Group II yı = 9.26 5.14 S 1.45 S2-2.81
Expert Solution
Step 1

This question is abut hypothesis testing independet sample t test. In this we have to test whether the aveage activity after lesion is less in hamsters.

Step 1)

Stating null and alternative hypothesis

In the null hypothesis we will assume that average of Group1 which is our control Group is equal to Group 2 which is our lesion group.

And in alternative hypothesis we will sate that average of Group2 which is our lesion Group is less than average of Group1.

We used less than symbol in alternative hypothesis because we have to test that did the lesion significantly reduce the activity of hamsters so reduce word leads us to use less than symbol.

Statistics homework question answer, Step 1, Image 1
Step 2

Step 2)

Getting the Critical value from t table

Since we have not been given the population standard deviation in question so we will use the t table and will find the critical value from the t table

In order to find the critical value we need to see three things

Degree of freedom which is n1 + n2 – 2 = 15+12-2 = 25

Alpha level or significance level : this is given in question to be 0.05

Tails of test: Its one tailed test (left tailed test) because if we have either less than symbol or greater than symbol in alternative hypothesis so it will be called one tailed test since it has less symbol so it will be called left tailed test.

Critical value will be = 1.708

So we get the critical value of 1.708 but since the test is left tailed test so the critical value will be -1.708

Statistics homework question answer, Step 2, Image 1
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