Question

Can you please explain every question step by step. I am trying to understand it as best as I can.

Dr. Smith believes that exposing babies and children to classical music with raise their intelligence quotient (IQ). To test this idea, he administers IQ tests to 10 children of his friends who listen to classical music on a regular basis. He then compares the results from these 10 kids to the national average, which is 100. The results are below:

T-Test

One-Sample Statistics

N Mean Std. Deviation Std. Error Mean IQ

10 113.5000 14.3469 4.5369 One-Sample Test

Test Value = 100

95% Confidence Interval of the Difference t df Sig. (2-tailed) Mean Difference Low er Upper IQ 2.976 9 .016 13.5000 3.2369 23.7631

a. What is the dependent variable?

b. What is the null hypothesis?

c. What is the alternative hypothesis?

d. What are the degrees of freedom?

e. What is t-obtained?

f. What is the 2-tailed p-value?

g. What is the 1-tailed p-value?

h. Should the null hypothesis be rejected? Why, or why not?

Answer 1

1. The depedent variable is Intelligence quotient (IQ) of children.
2. Null hypothesis H₀: μ = 100
3. Alternative hypothesis H₁: μ ≠100
4. The degrees of freedom are n-1 = 9
5. The t obtained Test statistics: t = (xbar – μ) / (s/√n) = =(113.5 - 100)/(14.3469/SQRT(10))= 2.976

   Test statistics = 2.976

6. The 2-tailed p-value = 2*P( t > 2.976) = 0.016
7. The 1 – tailed p – value = P( t > 2.976) = 0.008
8. If the 2-tailed p-value = 0.016 < 0.05, the null hypothesis is rejected at 0.05 level of significance.
   If the 2-tailed p-value = 0.016 > 0.01, the null hypothesis is not rejected at 0.01 level of significance.