Small sample test concerning 2 means with different variances

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In summary, the article discusses the growing concern of deteriorating municipal pipeline networks and states that the fusion process can increase the average tensile strength. Data on tensile strength of linen specimens with and without the fusion process are provided, and a test is carried out to determine if the data supports the conclusion. The results show that the data does support the conclusion, despite the assumption that the populations are normal.
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toothpaste666
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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
 

1. What is a small sample test concerning 2 means with different variances?

A small sample test concerning 2 means with different variances is a statistical test used to compare the means of two independent groups when the sample sizes are small and the variances of the two groups are different.

2. When is a small sample test concerning 2 means with different variances used?

This test is typically used when the assumptions for a standard t-test are not met, such as when the sample sizes are less than 30 or the variances of the two groups are not equal.

3. How is a small sample test concerning 2 means with different variances conducted?

The test involves calculating a test statistic, usually the Welch's t-test, and comparing it to a critical value from a t-distribution. The critical value is determined based on the degrees of freedom, which takes into account the sample sizes and variances of the two groups.

4. What is the difference between a small sample test concerning 2 means with different variances and a standard t-test?

A standard t-test assumes that the sample sizes are large and the variances of the two groups are equal. A small sample test concerning 2 means with different variances relaxes these assumptions and is more suitable for analyzing data from smaller sample sizes or when the variances are unequal.

5. What are the advantages of using a small sample test concerning 2 means with different variances?

This test allows for a more accurate comparison of the means of two groups when the sample sizes are small and the variances are different. It also does not require the data to follow a normal distribution, making it more robust for non-normal data.

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