A statistics teacher wants to see if there is any difference in the abilities of
ID: 3320537 • Letter: A
Question
A statistics teacher wants to see if there is any difference in the abilities of students enrolled in statistics today (2016) and those enrolled five years ago (2011). A sample of final examination scores from students enrolled today and from students enrolled five years ago was taken. You are given the following information:
2016 2011
the sample mean score (xbar) 88 82
the population variance (sigma squared) 54 112.5
the sample size (n) 36 45
Fill in Multiple Blanks:
The point estimate for the difference between the means of the two populations (that is, mu 2016 - mu 2011) is . In your answer, show one (1) digit to the right of the decimal point, for example, 1.0, 1.2. Apply the appropriate rounding rule if necessary.
The standard error of (xbar 2016 - xbar 2011) is . In your answer, show one (1) digit to the right of the decimal point, for example, 1.0, 1.2. Apply the appropriate rounding rule if necessary.
The 95% confidence interval for the difference between the two population means is from to . In your answer, show one (1) digit to the right of the decimal point, for example, 1.0, 1.2. Apply the appropriate rounding rule if necessary.
The statistics teacher wants to test the following set of hypotheses:
Ho: mu 2016 - mu 2011 = 0
Ha: mu 2016 - mu 2011 /= 0.
The actual value of the test statistic for the difference between the two population means is . In your answer, show one (1) digit to the right of the decimal point, for example, 1.0, 1.2. Apply the appropriate rounding rule if necessary.
Since the actual value of the test statistic is (greater, less) than the critical value of the test statistic, the teacher (rejects, does not reject) the null hypothesis and concludes that there (is, is no) statistically significant difference in the average final examination scores between the class of 2016 and the class of 2011.
Explanation / Answer
Q1.
TRADITIONAL METHOD
given that,
mean(x)=88
standard deviation , 1 =7.3484
population size(n1)=36
y(mean)=82
standard deviation, 2 =10.6066
population size(n2)=45
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((53.99898/36)+(112.49996/45))
= 2
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
since our test is two-tailed
value of z table is 1.96
margin of error = 1.96 * 2
= 3.92
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (88-82) ± 3.92 ]
= [2.08 , 9.92]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=88
standard deviation , 1 =7.3484
number(n1)=36
y(mean)=82
standard deviation, 2 =10.6066
number(n2)=45
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 = [ ( 88-82) ±Z a/2 * Sqrt( 53.99898256/36+112.49996356/45)]
= [ (6) ± Z a/2 * Sqrt( 4) ]
= [ (6) ± 1.96 * Sqrt( 4) ]
= [2.08 , 9.92]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95% sure that the interval [2.08 , 9.92] 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
Q2.
Given that,
mean(x)=88
standard deviation , 1 =7.3484
number(n1)=36
y(mean)=82
standard deviation, 2 =10.6066
number(n2)=45
null, Ho: u1 = u2
alternate, H1: 1 != u2
level of significance, = 0.05
from standard normal table, two tailed z /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=88-82/sqrt((53.99898/36)+(112.49996/45))
zo =3
| zo | =3
critical value
the value of |z | at los 0.05% is 1.96
we got |zo | =3 & | z | =1.96
make decision
hence value of | zo | > | z | and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 3 ) = 0.0027
hence value of p0.05 > 0.0027,here we reject Ho
ANSWERS
---------------
null, Ho: u1 - u2 = 0
alternate, H1: 1 - u2 != 0
test statistic: 3
critical value: -1.96 , 1.96
decision: reject Ho
p-value: 0.0027
statistically significant difference in the average final examination scores between the class of 2016 and the class of 2011.
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