Assume = 110, x_2 - 120, $_1 =8, 5, -12, n_1 = 10, and n_2 = 10. Find a 95% conf

Question

Assume = 110, x_2 - 120, $_1 =8, 5, -12, n_1 = 10, and n_2 = 10. Find a 95% confidence interval for the difference In the corresponding values of mu using the second approximation for degrees of freedom. The 95% confidence interval for the second approximation for degrees of freedom is (___, ______) (Give your answers to 4 decimal places). Assume X_1 = 110, X_2 =120, s_1 = 8, = 12, = 10, and n_2 = 10. Find a 95% confidence interval for the difference in the corresponding values of .u using the second approximation for degrees of freedom. Would you reject the null hypothesis that the population means are equal in favor of the two-sided alternative at significance level 0.05? Explain. We reject the null hypothesis in favor of the two-sided alternative at significance level 0.05, because the 95% confidence interval does not include 0. We cannot reject the null hypothesis in favor of the two-sided alternative at the 0.05 level, because 0 does not fall inside the 95% confidence interval. We do not reject the null hypothesis in favor of the two-sided alternative at the 0.05 level, because 0 falls in the 95% confidence interval. We reject the null hypothesis in favor of the two-sided alternative at the 0.05 level, because the P-value for the test is greater than 0.05.

Explanation / Answer

STEP 1:
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)=110
Standard deviation( sd1 )=8
Sample Size(n1)=10
Mean(x2)=120
Standard deviation( sd2 )=12
Sample Size(n2)=10
CI = [ ( 110-120) ±t a/2 * Sqrt( 64/10+144/10)]
= [ (-10) ± t a/2 * Sqrt( 20.8) ]
= [ (-10) ± 2.262 * Sqrt( 20.8) ]
= [-20.3163 , 0.3163]

STEP 2:
Given that,
mean(x)=110
standard deviation , s.d1=8
number(n1)=10
y(mean)=120
standard deviation, s.d2 =12
number(n2)=10
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.101
since our test is two-tailed
reject Ho, if to < -2.101 OR if to > 2.101
calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
s^2 = (9*64 + 9*144) / (20- 2 )
s^2 = 104
we use test statistic (t) = (x-y)/sqrt(s^2(1/n1+1/n2))
to=110-120/sqrt((104( 1 /10+ 1/10 ))
to=-10/4.5607
to=-2.1926
| to | =2.1926
critical value
the value of |t | with (n1+n2-2) i.e 18 d.f is 2.101
we got |to| = 2.1926 & | t | = 2.101
make decision
hence value of | to | > | t | and here we reject Ho
p-value: two tailed ( double the one tail ) - ha : ( p != -2.1926 ) = 0.041
hence value of p0.05 > 0.041,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: -2.1926
critical value: -2.101 , 2.101
decision: reject Ho
p-value: 0.041

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