A 18 gram sample of a substance that's used to detect explosives has a k value of0.125
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Answer:
Step-by-step explanation:
Hello!
The manufacturer suspects that the time a part for cell phones is allowed to cool in the mold influences the occurrence of a defect in the finished housing.
The claim is that the longer the part is left to cool down, the fewer defects appear.
To test this claim, two random samples of 20 parts each were taken, one was left to cool down 10sec and the other was left to cool down for 20sec. After the cooling time, each one was visually inspected and scored based on their appearance from 1 (lowest) to 10 (highest).
Resulting:
X₁: Score given to a manufactured part after a cooling time of 10 seconds.
n₁= 20 parts
X[bar]₁= 3.35
S₁= 2.01
X₂: Score given to a manufactured part after a cooling time of 20 seconds.
n₂= 20 parts
X[bar]₂= 6.50
S₂= 1.54
Assuming both variables have a normal distribution, and the population variances are equal, the statistic to use is a student t for two independent samples with pooled sample standard deviation.
If the longer the cooling time, the fewer defects appear in the part, then the mean score will be greater for the sample with higher cooling time.
The statistical hypotheses are:
H₀: μ₁ ≥ μ₂
H₁: μ₁ < μ₂
α: 0.05
Sa= 1.95
This test is one tailed to the left and so is the p-value, you can reach the value as follow:
p-value: ≅ 0.00001
p-value < 0.00001
The p.value is less than the significance level, the decision is to reject the null hypothesis.
Using a level of significance of 5% there is enough statistical evidence to reject the null hypothesis, so you can say that the average inspection scores of the parts that cooled down for 10 seconds are less than the average inspections scores of the parts that cooled down for 20 seconds, which means that the longer the cooling time, the fewer defects appear in the cellphone housing.
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95% CI for the difference of means μ₁ - μ₂
The formula for the confidence interval is:
[(X[bar]₁-X[bar]₂) ± ]
[(3.35-6.50)±2.024*(1.95*)]
[-4.40;-1.90]
With a confidence level of 95%, you'd expect that the true value of the difference between the inspection scores of the parts that cooled down for 10 seconds and inspections scores of the parts that cooled down for 20 seconds will be included in the interval [-4.40;-1.90].
I hope it helps!