Independent random samples from normal populations produced the results shown in
ID: 3130975 • Letter: I
Question
Independent random samples from normal populations produced the results shown in the table to the right. Complete parts a through d. Calculate the pooled estimate of sigma^2. (Round to four decimal places as needed.) Do the data provide sufficient evidence to indicate that mu_2 >mu_1? Test using alpha = 0.10. No Yes Find a 95% confidence interval for (mu_1 - mu_2). The confidence interval is (Round to two decimal places as needed.) Which of the two inferential procedures, the test of hypothesis in part b or the confidence interval in part c: provides more information about (mu_1 - mu_2)? The confidence interval in part c provides more information about ((mu_1 - mu_)2). The test of hypothesis in part b provides more information about (mu_1 - mu_2).Explanation / Answer
Set Up Hypothesis
Null, Ho: u1 > u2
Alternative, H1: u1 < u2
Test Statistic
X (Mean)=1.96; Standard Deviation (s.d1)=0.4827
Number(n1)=5
Y(Mean)=3.65; Standard Deviation(s.d2)=0.1732
Number(n2)=4
Value Pooled variance S^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
S^2 = (4*0.233 + 3*0.03) / (9- 2 )
S^2 = 0.146
we use Test Statistic (t) = (X-Y)/Sqrt(S^2(1/n1+1/n2))
to=1.96-3.65/Sqrt((0.146( 1 /5+ 1/4 ))
to=-1.69/0.2563
to=-6.5933
| to | =6.5933
Critical Value
The Value of |t | with (n1+n2-2) i.e 7 d.f is 1.415
We got |to| = 6.5933 & | t | = 1.415
Make Decision
Hence Value of | to | > | t | and Here we Reject Ho
P-Value: Left Tail - Ha : ( P < -6.5933 ) = 0.00015
Hence Value of P0.1 > 0.00015,Here we Reject Ho
[ANSWER]
a. S^2 = 0.146
b. Yes, we have evidence
c) WHEN SD ARE EQUAL
CI = (x1 - x2) ± t a/2 * S^2 * Sqrt ( 1 / n1 + 1 / 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
Value Pooled variance S^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
S^2 = (4*0.233 + 3*0.03) / (9- 2 )
S^2 = 0.146
S = 0.382
t a/2 = n1+n2-2 = 7 d.f is 1.415
CI = [ ( 1.96-3.65 ) ± t a/2 * S * Sqrt( 1/5 + 1/4)]
= [ (-3.2) ± t a/2 * 0.382 * Sqrt( 1 /5+ 1/4 ) ]
= [ (-3.2) ± (1.415 * 0.382 * Sqrt( 1 /5+ 1/4 )) ]
= [-3.562, -2.8374] ~ [-3.56, -2.84]
c. [-3.56, -2.84]
d. the confidence interval in part c provides more informatoin about u1-u2