Drop down 1: do not reject, reject Drop down 2: extremely wrong, no, weak, very
ID: 3260729 • Letter: D
Question
Drop down 1: do not reject, reject
Drop down 2: extremely wrong, no, weak, very strong, strong
Suppose we have taken independent, random samples of sizes n_1 = 6 and n_2 = 6 from two normally distributed populations having means mu_1 and mu_2, and suppose we obtain x_1 = 266, x_2 = 215, s_1 = 4, s_2 = 6. Use critical values to test the null hypothesis H_0: mu_1 - mu_2 lessthanorequalto 40 versus the alternative hypothesis H_a: mu_1 - mu_2 > 40 by setting a equal to 10, 05, 01 and 001. Using the equal variance procedure, how much evidence is there that the difference between mu_1 and mu_2 exceeds 40? (Round your answer to 3 decimal places.) t = (Click to select) H_0 at alpha = 0.1, 0.05, and, 0.01, (Click to select) evidence.Explanation / Answer
i] t = 3.74 and p-value = 0.002 when alpha = 0.1
ii] t = 3.74 and p-value = 0.002 when alpha = 0.05
iii] t = 3.74 and p-value = 0.002 when alpha = 0.01
find these values by using following methods in MINITAB. Repeat this method for various values of alpha, we get same result only lower limit of confidence interval will change.
here only alpha = 0.001 is only case where p-value is greater than alpha.
in this case we accept Ho at alpha = 0.001. otherwise we reject null hypothesis.
Choose Stat > Basic Statistics > 2-Sample T.
Choose Summarized data.
Enter first sample and second sample data.
Check Assume equal variances. Click OK.
then choose confidence level = 1 - alpha%
Check alternative then click ok
Session window output
MTB > TwoT 6 266 4 6 215 6;
SUBC> Pooled;
SUBC> Confidence 90;
SUBC> Test 40;
SUBC> Alternative 1.
Two-Sample T-Test and CI
Sample N Mean StDev SE Mean
1 6 266.00 4.00 1.6
2 6 215.00 6.00 2.4
Difference = mu (1) - mu (2)
Estimate for difference: 51.00
90% lower bound for difference: 46.96
T-Test of difference = 40 (vs >): T-Value = 3.74 P-Value = 0.002 DF = 10
Both use Pooled StDev = 5.0990
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