In: Statistics and Probability
all hypothesis testing problems must include the null and alternative hypotheses and report the p-value of the data.
A random sample of 45 students took an SAT preparation course
prior to taking the SAT. The sample mean of their quantitative SAT
scores was 575 with a s.d. of 90, and the sample mean of their
verbal SAT scores was 530 with a s.d. of 110.
a) Construct 95% confidence intervals for the mean
quantitative SAT and the mean verbal SAT scores of all students who
take this course.
b) Construct 95% confidence intervals for the
standard deviations of the QSAT and VSAT scores of all students who
take this course.
c) What sample size would be needed to estimate
the mean VSAT score with 95% confidence and with error of no more
than 5 if it is assumed that the s.d. is no more than 110?
a) Suppose the mean scores for all students who
took the SAT at that time was 535 for the quantitative and 505 for
the verbal? Do the means for students who take this course differ
from the means for all students at the 5% level of
significance?
a)The 95% confidence interval for the mean quantitative SAT score is
where ,s = 90 , n=45
For 95% confidence with 44 df
therefore 95% confidence interval is
=
= ( 547.96 , 602.04)
The 95% confidence interval for the mean verbal SAT score is
where ,s = 110 , n=45
For 95% confidence with 44 df
therefore 95% confidence interval is
=
= ( 514.91 , 608.05)
b) 95% confidence interval for standard deviation of QSAT
s= 90, n=45, df=44
threfore 95% confidence interval for standard deviation of QSAT is
(74.51 , 113.69)
95% confidence interval for standard deviation of VSAT
s= 110, n=45, df=44
threfore 95% confidence interval for standard deviation of QSAT is
(91.06 , 138.95)
c)given
margin of error = 5
therefore n = 1859
sample size neede = 1859
d) QSAT
The null and alternative hypothesis
Test statistic
=
= 2.98
df =44
P value = 0.0047
Since P value < 0.05
we reject H0
At 5% level there is sufficient evidence to conclude that QSAT scores of students who take this course differ from means of all students .
VSAT
The null and alternative hypothesis
Test statistic
=
= 1.52
df =44
P value = 0.1357
Since P value > 0.05
we fail to reject H0
At 5% level there is not sufficient evidence to conclude that VSAT scores of students who take this course differ from means of all students .