Question

The subjects were 30 male and 30 female right-handed introductory psychology students who volunteered to parti cipate in exchange for course credit. Initial testing on spatial and verbal tests revealed the following summary statistics. Note that the scores on the spatial task can range from 0 to 40, whereas those on the verbal task can go from 0 to 20. The distributions are not strongly skewed on either scale or for males or females Spatial Verbal Sex Mean 10.20 7.80 SD 2.70 2.50 Mean 10.90 15.10 SD 3.00 3.40 Male Female a. Consider comparing males to females with regard to performance on the spatial assessment task. State the appropriate null and alternative hypotheses in the context of the study b. Explain why it is valid to use the theory-based method for producing ap-value to test the hypotheses stated in part (a). c. Carry out the appropriate test to produce a p-value to test the hypotheses stated in part (a) and interpret the p-value d. Find a 95% confidence interval for the difference in mean scores of males and females with regard to performance on spatial assessments. Interpret the interval. e. Based on your p-value, state a conclusion in the context of the study. Be sure to comment on statistical signifi cance, estimation (confidence interval), causation, and generalization. f. Repeat the investigation comparing males and females, this time on verbal performance. Be sure to address the questions asked in parts (a)-(e)

Answer 1

Given that,
mean(x)=10.2
standard deviation , sigma1 =2.7
number(n1)=30
y(mean)=7.8
standard deviation, sigma2 =2.5
number(n2)=30
null, Ho: u1 = u2
alternate, H1: μ1 != u2
level of significance, alpha = 0.05
from standard normal table, two tailed z alpha/2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic (z) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
zo=10.2-7.8/sqrt((7.29/30)+(6.25/30))
zo =3.572
| zo | =3.572
critical value
the value of |z alpha| at los 0.05% is 1.96
we got |zo | =3.572 & | z alpha | =1.96
make decision
hence value of | zo | > | z alpha| and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 3.572 ) = 0.00035
hence value of p0.05 > 0.00035,here we reject Ho
ANSWERS
---------------
a.
null, Ho: u1 = u2
alternate, H1: μ1 != u2
c.
test statistic: 3.572
critical value: -1.96 , 1.96
b.
p-value: 0.00035
decision: reject Ho
d.
TRADITIONAL METHOD
given that,
mean(x)=10.2
standard deviation , σ1 =2.7
population size(n1)=30
y(mean)=7.8
standard deviation, σ2 =2.5
population size(n2)=30
I.
stanadard error = sqrt(s.d1^2/n1)+(s.d2^2/n2)
where,
sd1, sd2 = standard deviation of both
n1, n2 = sample size
stanadard error = sqrt((7.29/30)+(6.25/30))
= 0.6718
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
since our test is two-tailed
value of z table is 1.96
margin of error = 1.96 * 0.6718
= 1.3168
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (10.2-7.8) ± 1.3168 ]
= [1.0832 , 3.7168]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=10.2
standard deviation , σ1 =2.7
number(n1)=30
y(mean)=7.8
standard deviation, σ2 =2.5
number(n2)=30
CI = x1 - x2 ± Z a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
where,
x1,x2 = mean of populations
sd1,sd2 = standard deviations
n1,n2 = size of both
a = 1 - (confidence Level/100)
Za/2 = Z-table value
CI = confidence interval
CI = [ ( 10.2-7.8) ±Z a/2 * Sqrt( 7.29/30+6.25/30)]
= [ (2.4) ± Z a/2 * Sqrt( 0.4513) ]
= [ (2.4) ± 1.96 * Sqrt( 0.4513) ]
= [1.0832 , 3.7168]
-----------------------------------------------------------------------------------------------
e.
interpretations:
1. we are 95% sure that the interval [1.0832 , 3.7168] contains the difference between
true population mean U1 - U2
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the difference between
true population mean U1 - U2
3. Since this Cl does contain a zero we can conclude at 0.05 true mean
difference is zero.
conclusion:
we have enough evidence to support the claim that difference of means males and females of spatial assesement.
f.
Given that,
mean(x)=10.9
standard deviation , sigma1 =3
number(n1)=30
y(mean)=15.1
standard deviation, sigma2 =3.4
number(n2)=30
null, Ho: u1 = u2
alternate, H1: μ1 != u2
level of significance, alpha = 0.05
from standard normal table, two tailed z alpha/2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic (z) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
zo=10.9-15.1/sqrt((9/30)+(11.56/30))
zo =-5.073
| zo | =5.073
critical value
the value of |z alpha| at los 0.05% is 1.96
we got |zo | =5.073 & | z alpha | =1.96
make decision
hence value of | zo | > | z alpha| and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -5.073 ) = 0
hence value of p0.05 > 0,here we reject Ho
ANSWERS
---------------
i.
null, Ho: u1 = u2
alternate, H1: μ1 != u2
iii.
test statistic: -5.073
critical value: -1.96 , 1.96
decision: reject Ho
ii.
p-value: 0
iv.
TRADITIONAL METHOD
given that,
mean(x)=10.9
standard deviation , σ1 =3
population size(n1)=30
y(mean)=15.1
standard deviation, σ2 =3.4
population size(n2)=30
I.
stanadard error = sqrt(s.d1^2/n1)+(s.d2^2/n2)
where,
sd1, sd2 = standard deviation of both
n1, n2 = sample size
stanadard error = sqrt((9/30)+(11.56/30))
= 0.8278
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
since our test is two-tailed
value of z table is 1.96
margin of error = 1.96 * 0.8278
= 1.6226
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (10.9-15.1) ± 1.6226 ]
= [-5.8226 , -2.5774]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=10.9
standard deviation , σ1 =3
number(n1)=30
y(mean)=15.1
standard deviation, σ2 =3.4
number(n2)=30
CI = x1 - x2 ± Z a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
where,
x1,x2 = mean of populations
sd1,sd2 = standard deviations
n1,n2 = size of both
a = 1 - (confidence Level/100)
Za/2 = Z-table value
CI = confidence interval
CI = [ ( 10.9-15.1) ±Z a/2 * Sqrt( 9/30+11.56/30)]
= [ (-4.2) ± Z a/2 * Sqrt( 0.6853) ]
= [ (-4.2) ± 1.96 * Sqrt( 0.6853) ]
= [-5.8226 , -2.5774]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95% sure that the interval [-5.8226 , -2.5774] contains the difference between
true population mean U1 - U2
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the difference between
true population mean U1 - U2
3. Since this Cl does contain a zero we can conclude at 0.05 true mean
difference is zero

v.

conclusion:
we have enough evidence to support the claim that difference of means of males and females of verbal assesement