Small sample test concerning 2 means with different variances

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The discussion revolves around testing the difference in tensile strength between two groups: non-fused and fused linen specimens. Using a significance level of α = 0.05 and assuming unequal variances, the null hypothesis (H0) states that there is no difference in means, while the alternative hypothesis (H1) suggests that the mean tensile strength of non-fused specimens is less than that of fused specimens. The calculated t' value of -1.8 is less than the critical value of -1.746, leading to the rejection of H0, indicating that the data supports the conclusion that the fusion process increases tensile strength. There is some confusion regarding the assumption of normality for the populations, but the analysis appears to be correct based on the provided data. Overall, the findings suggest a significant difference in tensile strength due to the fusion process.
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Homework Statement


The deterioration of many municipal pipeline networks across the county is a growing concern. An article stated that the fusion process increased the average tensile strength. Data on tensile strength (psi) of linen specimens when a certain fusion process was used and when this process was not given are provided. The data are given below.

Tensile strength (psi)
Nofusion 2748 2700 2655 2822 2511 3149 3257 3123 3220 2753
Fused 3027 3356 3359 3297 3125 2910 2889 2902

Carry out a test to see whether the data support this conclusion. Use α = 0.05. Assume
σ1 ≠ σ2

n1 = 10 x1 = 2902.8 s1 = 277.3 n2 = 8 x2 = 3108.1 s2 = 205.9

The Attempt at a Solution


μ1 = nonfusion μ2 = fused
H0: μ1 - μ2 = 0
H1: μ1 - μ2 < 0

for a small sample with unequal standard devations we use t'
reject H0 when t' < -tα = -t.05 for the estimated degrees of freedom v where

v = (s1^2/n1 + s2^2/n2)^2/[(s1^2/n1)^2/(n1-1) + (s2^2/n2)^2/(n2-1)]
= (277.3^2/10 + 205.9^2/8)^2/[(277.3^2/10)^2/9 + (205.9^2/8)^2/7]
= 15.9 ≅ 16

so reject H0 when t' < -1.746

t' = [(x1 - x2) - 0]/sqrt(s1^2/n1 + s2^2/n2) = (2902.8 - 3108.1)/sqrt(277.3^2/10 + 205.9^2/8)
= (-205.3)/113.97 = - 1.8
t' = -1.8 < -1.746
we must reject H0. the data supports the conclusion.

is this correct? I am slightly confused about if i should use t' or not because the problem never states that the populations are normal
 
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toothpaste666 said:
is this correct?
Seems OK to me (I am not a statistician, just a mathematician that has read up on basic statistics).
 
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Thank you. hopefully the assumption that the populations are normal doesn't cause me too much trouble
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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