When the same 453 selected QC students are organized into two groups: male and f

Question

When the same 453 selected QC students are organized into two groups: male and female. The summary of the test scores are provided as follows: Std Deviation 110 Average score 203 250 Male 1015 100 Female c. What do these data suggest about the difference in gender-based performance for d. Is there a statistically significant difference in test performance between male and QC students? (1pt) female students? (4pts). In order to answer this question, you should: i) State your null and alternative hypothesis; ii) Compute the relevant t-statistic; iii) Obtain the p-value iv) State your conclusion at 5% significance level, and comment.

Explanation / Answer

Given that,
mean(x)=1010
standard deviation , s.d1=110
number(n1)=203
y(mean)=1015
standard deviation, s.d2 =100
number(n2)=250
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.05
from standard normal table, two tailed t /2 =1.972
since our test is two-tailed
reject Ho, if to < -1.972 OR if to > 1.972
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =1010-1015/sqrt((12100/203)+(10000/250))
to =-0.501
| to | =0.501
critical value
the value of |t | with min (n1-1, n2-1) i.e 202 d.f is 1.972
we got |to| = 0.50099 & | t | = 1.972
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -0.501 ) = 0.617
hence value of p0.05 < 0.617,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: -0.501
critical value: -1.972 , 1.972
decision: do not reject Ho
p-value: 0.617

we don't have evidence to support difference in test performance between male an
female students

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.