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Answer:
Step-by-step explanation:
Hello!
The objective of this experiment is to compare two compounds, designed to reduce braking distance, used in tire manufacturing to prove if the braking distance of SUV's equipped with tires made with compound 1 is shorter than the braking distance of SUV's equipped with tires made with compound 2.
So you have 2 independent populations, SUV's equipped with tires made using compound 1 and SUV's equipped with tires made using compound 2.
Two samples of 81 braking tests are made and the braking distance was measured each time, the study variables are determined as:
X₁: Braking distance of an SUV equipped with tires made with compound one.
Its sample mean is X[bar]₁= 69 feet
And the Standard deviation S₁= 10.4 feet
X₂: Braking distance of an SUV equipped with tires made with compound two.
Its sample mean is X[bar]₂= 71 feet
And the Standard deviation S₂= 7.6 feet
We don't have any information on the distribution of the study variables, nor the sample data to test it, but since both sample sizes are large enough n₁ and n₂ ≥ 30 we can apply the central limit theorem and approximate the distribution of both variables sample means to normal.
The researcher's hypothesis, as mentioned before, is that the braking distance using compound one is less than the distance obtained using compound 2, symbolically: μ₁ < μ₂
The statistical hypotheses are:
H₀: μ₁ ≥ μ₂
H₁: μ₁ < μ₂
α: 0.05
The statistic to use to compare these two populations is a pooled Z test
Z ≈ N(0;1)
The rejection region if this hypothesis test is one-tailed to the right, so you'll reject the null hypothesis to small values of the statistic. The critical value for this test is:
Decision rule:
If > -1.648 , then you do not reject the null hypothesis.
If ≤ -1.648 , then you reject the null hypothesis.
Since the statistic value is greater than the critical value, the decision is to not reject the null hypothesis.
At a 5% significance level, you can conclude that the average braking distance of SUV's equipped with tires manufactured used compound 1 is greater than the average braking distance of SUV's equipped with tires manufactured used compound 2.
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If inspection department wants to estimate the mean amount with 95% confidence level with standard deviation 0.05 then it needed a sample size of 97.
Given 95% confidence level, standard deviation=0.05.
We know that margin of error is the range of values below and above the sample statistic in a confidence interval.
We assume that the values follow normal distribution. Normal distribution is a probability that is symmetric about the mean showing the data near the mean are more frequent in occurence than data far from mean.
We know that margin of error for a confidence interval is given by:
Me=
α=1-0.95=0.05
α/2=0.025
z with α/2=1.96 (using normal distribution table)
Solving for n using formula of margin of error.
n=
=96.4
By rounding off we will get 97.
Hence the sample size required will be 97.
Learn more about standard deviation at brainly.com/question/475676
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The given question is incomplete and the full question is as under:
If the inspection division of a county weights and measures department wants to estimate the mean amount of soft drink fill in 2 liters bottles to within (0.01 liter with 95% confidence and also assumes that standard deviation is 0.05 liter. What is the sample size needed?