Question 1181620: One manufacturer claims that the average tensile strength of yarn A is less than the average tensile strength of yarn B by less than 12 kilograms.
To prove this claim, 50 pieces of type A yarn and 45 pieces of type B yarn were tested under similar conditions.
Type A yarn had an average tensile strength of 86.7 kilograms; while type B yarn had an average tensile strength of 77.8 kilograms.
Test the manufacturer's claim using a significance level of 0.005 knowing that, from previous studies, type A yarn has a standard deviation of 6.28 kilograms while type B yarn has a standard deviation of 5.61 kilograms.
a)Solution:
b)Parameter under study:
c)Planteamiento de hipótesis:
d)Nivel de significancia
e)Statistical:
f)Rejection region:
g)Decision making:
h)Conclution:
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Here's a complete hypothesis test for the manufacturer's claim:
**a) Solution:**
We will use a two-sample z-test for the difference between means since the sample sizes are large (n > 30) and the population standard deviations are known.
**b) Parameter under study:**
The parameter under study is the difference between the average tensile strength of yarn A (μA) and the average tensile strength of yarn B (μB). We are interested in whether μA - μB < 12.
**c) Statement of Hypotheses:**
* Null hypothesis (H₀): μA - μB ≥ 12 (The difference is greater than or equal to 12 kg)
* Alternative hypothesis (H₁ or Ha): μA - μB < 12 (The manufacturer's claim: The difference is less than 12 kg)
**d) Significance level:**
α = 0.005
**e) Test statistic:**
The test statistic for a two-sample z-test is:
z = [(xA - xB) - (μA - μB)₀] / sqrt(σA²/nA + σB²/nB)
Where:
* xA and xB are the sample means
* (μA - μB)₀ is the hypothesized difference under H₀ (12)
* σA and σB are the population standard deviations
* nA and nB are the sample sizes
z = [(86.7 - 77.8) - 12] / sqrt((6.28²/50) + (5.61²/45))
z = [-3.1] / sqrt(0.788 + 0.700)
z = -3.1 / sqrt(1.488)
z ≈ -3.1 / 1.22
z ≈ -2.54
**f) Rejection region:**
Since this is a one-tailed test (less than), we look for the critical z-value that corresponds to α = 0.005 in the left tail of the standard normal distribution. Using a z-table or calculator, we find that the critical z-value is approximately -2.576.
The rejection region is z < -2.576.
**g) Decision making:**
Our calculated test statistic (z ≈ -2.54) falls *within* the rejection region (z < -2.576) .
Therefore, we *reject* the null hypothesis.
**h) Conclusion:**
There is sufficient evidence at the 0.005 significance level to support the manufacturer's claim that the average tensile strength of yarn A is less than the average tensile strength of yarn B by less than 12 kilograms.
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