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nalin [4]
3 years ago
7

An airplane manufacturer needs to purchase an aluminum alloy for use in constructing a part for their new line of jumbo jets. It

is important that the alloy be very strong. Three suppliers, A, B and C can provide this alloy. In order to make their own comparison of the alloys produced by these suppliers they obtain samples from each one and test their tensile strength. The results follow:
A B C

200 psi* 177 psi 202 psi

180 167 212

190 188 214

190 180 188

190 180 188

182 * psi = pounds per square inch

What statistical test/procedure would you use to determine whether there is a significant difference in the tensile strength of the alloys produced by the three suppliers (a = .05)?

t-test

ANOVA

F-test of variances

Chi-Square
Mathematics
1 answer:
Nostrana [21]3 years ago
6 0

Answer:

Step-by-step explanation:

Hello!

In this example, you have 3 populations determined by each aluminum alloy A, B, and C and want to test if their tensile strength is the same or not. So the study variable is the tensile strength of the aluminum and it is tested in each sample taken of each alloy. The parameters of interest are the average tensile strength of each alloy.

The best statistical procedure to compare these three population is an ANOVA test of one factor(tensile strength) with three levels (alloy A, alloy B, and alloy C)

I hope it helps!

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Answer:

a) \bar X = \frac{2+2+1+3+1+0+4+1}{8}= 1.75

b) The margin of error indicates we can be 95%confident that the sample mean falls within 0.89 of the population mean

Step-by-step explanation:

Part a

The best point of estimate for the population mean is the sample mean given by:

\bar X = \frac{\sum_{i=1}^n X_i}{n}

Since is an unbiased estimator E(\bar X) = \mu

Data given: 2 , 2 , 1 , 3 , 1 , 0 , 4 , 1

So for this case the sample mean would be:

\bar X = \frac{2+2+1+3+1+0+4+1}{8}= 1.75

Part b

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

\bar X represent the sample mean for the sample  

\mu population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

The margin of error is given by this formula:

ME=t_{\alpha/2}\frac{s}{\sqrt{n}}    (2)

And for this case we know that ME =0.89 with a confidence of 95%

So then the limits for our confidence level are:

Lower= \bar X -ME= 1.75- 0.89=0.86

Upperr= \bar X +ME= 1.75+0.89=2.64

So then the best answer for this case would be:

The margin of error indicates we can be 95%confident that the sample mean falls within 0.89 of the population mean

7 0
3 years ago
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