A sample of 125 pieces of yarn had mean breaking strength 6.1 N and standard deviation 0.7 N. A new batch of yarn was made, using new raw materials from a different vendor. In a sample of 75 pieces of yarn from the new batch, the mean breaking strength was 5.8 N and the standard deviation was 1.0 N. Find a 90% confidence interval for the difference in mean breaking strength between the two types of yarn.

Respuesta :

Answer:

The 90% confidence interval for the difference in mean breaking strength between the two types of yarn is (0.08N, 0.52N).

Step-by-step explanation:

Before building the confidence interval, we need to understand the central limit theorem and subtraction of normal variables.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Subtraction of normal variables:

When we subtract two normal variables, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.

A sample of 125 pieces of yarn had mean breaking strength 6.1 N and standard deviation 0.7 N.

This means that [tex]\mu_1 = 6.1, s_1 = \frac{0.7}{\sqrt{125}} = 0.0626[/tex]

In a sample of 75 pieces of yarn from the new batch, the mean breaking strength was 5.8 N and the standard deviation was 1.0 N.

This means that [tex]\mu_2 = 5.8, s_2 = \frac{1}{\sqrt{75}} = 0.1155[/tex]

Distribution of the difference:

[tex]\mu = \mu_1 - \mu_2 = 6.1 - 5.8 = 0.3[/tex]

[tex]s = \sqrt{s_1^2+s_2^2} = \sqrt{0.0626^2+0.1155^2} = 0.1314[/tex]

Confidence interval:

We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:

[tex]\alpha = \frac{1 - 0.9}{2} = 0.05[/tex]

Now, we have to find z in the Ztable as such z has a pvalue of [tex]1 - \alpha[/tex].

That is z with a pvalue of [tex]1 - 0.05 = 0.95[/tex], so Z = 1.645.

Now, find the margin of error M as such

[tex]M = zs[/tex]

[tex]M = 1.645*0.1314 = 0.22[/tex]

The lower end of the interval is the sample mean subtracted by M. So it is 0.3 - 0.22 = 0.08N

The upper end of the interval is the sample mean added to M. So it is 0.3 + 0.22 = 0.52N

The 90% confidence interval for the difference in mean breaking strength between the two types of yarn is (0.08N, 0.52N).

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