Question 1207902: Independent random samples of n1 = 18 and n2 = 13 observations were selected from two normal populations with equal variances.
DATA:-
____________________Population (Ignore the lines)
____________________ 1_____2 (Ignore the lines)
Sample Size_________18____13 (Ignore the lines)
Sample Mean________34.6___32.1 (Ignore the lines)
Sample Variance_____4.5___5.9 (Ignore the lines)
(a) Find the rejection region for the test in part (a) for 𝛼 = 0.01. (If the test is one-tailed, enter NONE for the unused region. Round your answers to three decimal places.)
t > _______
t < _______
(b) Find the value of the test statistic. (Round your answer to three decimal places.)
t =
(c) Find the approximate p-value for the test.
p-value < 0.010
i) 0.010 < p-value < 0.020
ii) 0.020 < p-value < 0.050
iii) 0.050 < p-value < 0.100
iv) 0.100 < p-value < 0.200
v) p-value < 0.200
Answer by ElectricPavlov(122)   (Show Source):
You can put this solution on YOUR website! **1. Calculate the Pooled Variance**
* Pooled variance (s_p^2) = ((n1 - 1) * s1^2 + (n2 - 1) * s2^2) / (n1 + n2 - 2)
* s_p^2 = ((18 - 1) * 4.5 + (13 - 1) * 5.9) / (18 + 13 - 2)
* s_p^2 = 5.06
**2. Calculate the Degrees of Freedom**
* Degrees of freedom (df) = n1 + n2 - 2 = 18 + 13 - 2 = 29
**3. Find the Critical Value**
* For a two-tailed test with α = 0.01 and df = 29, the critical values are:
* t_critical = ±2.756
**4. Calculate the Test Statistic**
* t = (x̄1 - x̄2) / √(s_p^2 * (1/n1 + 1/n2))
* t = (34.6 - 32.1) / √(5.06 * (1/18 + 1/13))
* t = 3.048
**5. Determine the P-value**
* Using a t-distribution table or statistical software, find the p-value associated with the calculated t-statistic (3.048) and degrees of freedom (29).
* The p-value will be less than 0.01.
**Therefore:**
* **(a) Rejection Region:**
* t > 2.756
* t < -2.756
* **(b) Test Statistic (t): 3.048**
* **(c) P-value: p-value < 0.010**
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