1 There are two different grades Grade one sample size m129

1. There are two different grades.

Grade one: sample size m=129, sample mean (xbar)=107.6, sample StDev s1=1.3

Grade two: sample size n=129, sample mean (ybar)=123.6, sample StDev s2=2.0

a) What is the standard error of (xbar-ybar), i.e., se(xbar-ybar)

b) Calculate the 90% CI (u1-u2)

c) Interpret the above results

d) If the sample sizes are changed to m=12 and n=10, what is the standard error of (xbar-ybar), i.e., se(xbar-ybar)? Calculate the 99% CI of (u1-u2)

2) There are two different groups

Grade one: sample size m=4, sample mean xbar=13.90 and sample StDev s1=1.225

Grade two: sample size n=4, sample mean ybar=12.20 and sample StDev s2=1.010.

a) what is the pooled standard error of (xbar-ybar)

b) calculate the 95% pooled confidence interval for (u1-u2)

Solution

Q1.

a)
Stanadard error = Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
= Sqrt( 1.69/129+4/129)
= 0.21002
b)
CI = x1 - x2 ± t a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
Where,
x1 = Mean of Sample 1, x2 = Mean of sample2
sd1 = SD of Sample 1, sd2 = SD of sample2
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x1)=107.6
Standard deviation( sd1 )=1.3
Sample Size(n1)=129
Mean(x2)=123.6
Standard deviation( sd2 )=2
Sample Size(n1)=129
CI = [ ( 107.6-123.6) ±t a/2 * Sqrt( 1.69/129+4/129)]
= [ (-16) ± t a/2 * Sqrt( 0.04) ]
= [ (-16) ± 1.657 * Sqrt( 0.04) ]
= [-16.35 , -15.65]

c)
Stanadard error = Sqrt( 1.69/12+4/10) = 0.7354
d)
CI = [ ( 107.6-123.6) ±t a/2 * Sqrt( 1.69/12+4/10)]
= [ (-16) ± t a/2 * Sqrt( 0.54) ]
= [ (-16) ± 3.25 * Sqrt( 0.54) ]
= [-18.39 , -13.61]