Suppose we observe a random sample of five measurements: 10, 13, 15, 15, 17, from a normal distribution with unknown mean mu_1 and unknown variance sigma_1^2, A second random sample from another normal population with unknown mean mu_2 and unknown variance sigma_2^2 yields the measurements: 13, 7, 9, 11.

django0a6

django0a6

Answered question

2022-11-19

Getting P-value While Using Variance
Suppose we observe a random sample of five measurements: 10, 13, 15, 15, 17, from a normal distribution with unknown mean μ 1 and unknown variance σ 1 2 , A second random sample from another normal population with unknown mean μ 2 and unknown variance σ 2 2 yields the measurements: 13, 7, 9, 11.
a) Test for evidence that σ 1 > 1.0. Complete the P-value for this test as accurately as possible. Draw a conclusion at α = 0.05.
Here's what I've done so far:
Step 1: Calculate σ 1
σ 1 = ( 10 14 ) 2 + ( 13 14 ) 2 + ( 15 14 ) 2 + ( 15 14 ) 2 + ( 17 14 ) 2 5 = 28 5 = 2.366
Step 2: Set up Hypothesis Test
H 0 : σ 1 = 2.366
H a : σ 1 > 1.0
How do I proceed from here? Thanks.
EDIT:
Also have this question, and would appreciate some insight.
b) Use the pivotal method(and a pivotal statistic with F distribution) to derive a 95% confidence interval for σ 2 σ 1 . Work it out for these data. And test the null hypothesis that σ 2 = σ 1 at the 5% level of significance.

Answer & Explanation

retalibry9

retalibry9

Beginner2022-11-20Added 16 answers

Step 1
for part a), you've made a wrong null hypothesis test. You want to test for evidence that σ 1 > 1 σ 1 2 > 1 . Thus, your hypotheses should be:
H 0 : σ 1 2 = 1
H 1 : σ 1 2 > 1
We need n 1 which is 5. We also need our sample variance, which you have the square root of. Thus, S 2 = 5.6 and now we calculate our chi-square statistic, which is
χ 2 = ( n 1 ) s 2 σ 0 2 = ( 5 1 ) ( 5.6 ) 1 = 22.4
We know our α = .05, so we need to find χ α 2 ( n 1 ) = χ .05 2 ( 4 ) = 9.488.. Since 22.4 9.488, we reject H 0 . By p-value, we can see that at α = .01 , P ( W 13.28 ) = .01, so our p-value is less than .01. Also, P ( W 14.86 ) = .005, so it would appear that our p-value for this problem is VERY small indeed!
Step 2
for part b), we know that s 1 2 = 5.6 and s 2 2 = ( 13 10 ) 2 + ( 7 10 ) 2 + ( 9 10 ) 2 + ( 11 10 ) 2 4 = 5.. ALso, let m be the number of measurements from the first random sample and let n be the number of measurements from the second random sample. Considering the scope of deriving the confidence, I give it to you freely here:
.95 C I ( σ 2 2 σ 1 2 ) = [ 1 F α 2 ( n 1 ) ( m 1 ) s 2 2 s 1 2 , F α 2 ( m 1 ) ( n 1 ) s 2 2 s 1 2 ] = [ 1 F .025 ( 4 ) ( 3 ) 5.6 5 , F .025 ( 3 ) ( 4 ) 5.6 5 ]
Since I gave you the freebee, you should try and finish the hypothesis test for the ratio using what you know of the F-statistic and the information provided...

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