Provided below are summary statistics for independent simple random samples from

Question

Provided below are summary statistics for independent simple random samples from two normal populations. Use the pooled t-test and the pooled t-interval procedure to conduct the required hypothesis test and obtain the specified confidence interval. X_1 = 12, s_1 = 2.2, n_1 = 11, x_2 = 16, s_2 = 2.4, n_2 = 11 Two-tailed test, alpha = 0.10 90% confidence interval First, what are the correct hypotheses for a two-tailed test? H_0:mu_1 mu_2 H_a:mu_1 = mu_2 H_0:mu_1 = mu_2 H_a:mu_1 mu_2 H_a:mu_1 = mu_2 H_0:mu_1 mu_2 H_a:mu_1 mu_2 H_0:mu_1 = mu_2 H_a:mu_1 mu_2 H_0:mu_1 = mu_2 H_a:mu_1 > mu_2 H_0:mu_1

Explanation / Answer

Set Up Hypothesis
Null Hypothesis, There Is NoSignificance between them Ho: u1 = u2
Alternative Hypothesis, There Is Significance between them H1: u1 != u2
Test Statistic
X (Mean)=12; Standard Deviation (s.d1)=2.2
Number(n1)=11
Y(Mean)=16; Standard Deviation(s.d2)=2.4
Number(n2)=11
Value Pooled variance S^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
S^2 = (10*4.84 + 10*5.76) / (22- 2 )
S^2 = 5.3
we use Test Statistic (t) = (X-Y)/Sqrt(S^2(1/n1+1/n2))
to=12-16/Sqrt((5.3( 1 /11+ 1/11 ))
to=-4/0.9816
to=-4.0748
| to | =4.0748
Critical Value
The Value of |t | with (n1+n2-2) i.e 20 d.f is 1.725
We got |to| = 4.0748 & | t | = 1.725
Make Decision
Hence Value of | to | > | t | and Here we Reject Ho
P-Value: Two Tailed ( double the one tail ) - Ha : ( P != -4.0748 ) = 0.0005
Hence Value of P0.1 > 0.0005,Here we Reject Ho

[ANS]
1.Ho: u1 = u2, H1: u1 != u2
2. to=-4.0748
3. ( P != -4.0748 ) = 0.0005

CI = x1 - x2 ± t a/2 * Sqrt(S^2(1/n1+1/n2))
Where,
x1 = Mean of Sample 1, x2 = Mean of sample2
sd1 = SD of Sample 1, sd2 = SD of sample2
Value Pooled variance S^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x1)=12
Standard deviation( sd1 )=2.2
Sample Size(n1)=11
Mean(x2)=16
Standard deviation( sd2 )=2.4
Sample Size(n2)=11
S^2 = (10*4.84 + 10*5.76) / (22- 2 )
S^2 = 5.3
CI = [ ( 12-16) ± t a/2 * 2.302 Sqrt( 1/11+1/11)]
= [ (-4) ± t a/2 * 2.302 * Sqrt( 0.182) ]
= [ (-4) ± 2.001 * 2.302 * Sqrt( 0.182) ]
= [ (-5.964 , -2.036 ]

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