Question

Autism may be marked by different patterns of brain growth during early life. Researchers measured brain volume (in milliliters, using an MRI) in 29 boys who were later diagnosed with autism. Brain volume is also available for 13 boys who did not develop autism; these 13 boys serve as the control group. The sample statistics are presented for both groups in the table below. Assume that the population variances are equal. Use a sigificance level of 0.05 to test for a difference between the autism and control groups.

Groups	Sample size	Sample mean	Sample SD
Autism (group 1)	n1 = 29	y1 = 1364	s1 = 87.77
Control (group 2)	n2 = 13	y2 = 1244.63	s2 = 69.98

1. The null hypothesis for this test is:
H₀ : μ₁ − μ₂ > 0
H₀ : μ_d = 0
H₀ : μ_d1 − μ_d2 ≠ 0
H₀ : μ_d1 − μ_d2 > 0
H₀ : μ_d > 0
H₀ : μ₁ − μ₂ = 0

Answer Submitted: Your final submission will be graded when the time limit is reached.

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1. The alternative hypothesis for this test is:
H_A : μ_d > 0
H_A : μ₁ − μ₂ > 0
H_A : μ_d1 − μ_d2 > 0
H_A : μ_d ≠ 0
H_A : μ_d1 − μ_d2 = 0
H_A : μ₁ − μ₂ ≠ 0

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3. The standard error of the estimated difference between the autism and control groups is (give answer to 4 places after decimal):

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4. The value of the test statistic is (give the answer to 4 places after decimal):

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5. The P-value is approximately:
< 0.01
0.01 - 0.05
0.05 - 0.1
0.1 - 0.2
0.2 - 0.5
> 0.5

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6. According to the P-value method, what decision would you make?
Because P-value < α, I reject H₀.
Because P-value > α, I reject H₀.
Because P-value > α, I fail to reject H₀.
Because P-value < α, I fail to reject H₀.

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7. If you incorrectly reject the Null hypothesis, what type of error did you make?
None. If I correctly followed the rejection rule I would not make an error.
Type-II error
Type-I error

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8. The 90% confidence interval for the true difference between brain volumes is (give answers to two places after decimal):

lower bound:  upper bound:

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Answer 1

1) Null Hypothesis:

ANSWER: F.

H0 : μ1 − μ2 = 0

2) Alternative Hypothesis:

ANSWER: F

HA : μ1 − μ2 ≠ 0

3) Find: Standard error: Assume that the population variances are equal

4)

Test statistic:

5)

Degree of Freedom = n1+n2-2 = 29+13-2 = 40

P-value: 0.0001   .............Using t table

ANSWER: A.   < 0.01

6)

ANSWER: A.

Because P-value < α, I reject H0.