Question

Two types of plastics are suitable for use by an electronics component manufacturer. The breaking strength of these plastics is very important. From a random sample size of n₁=18, and n₂=16, we obtained that X₁-bar=151.2, S₁=1.4 and X₂-bar=152.3, S₂=1.65.

Calculate a 95% confidence interval on the ratio of variances.

With 95% confidence, what is the right-value of the two-sided confidence interval on the ratio of variances?

Your Answer:

Two types of plastics are suitable for use by an electronics component manufacturer. The breaking strength of these plastics is very important. From a random sample size of n₁=18, and n₂=16, we obtained that X₁-bar=151.2, S₁=1.4 and X₂-bar=152.3, S₂=1.65.

Calculate a 95% confidence interval on the difference in means, assuming that the population variances are equal.

With 95% confidence, what is the left-value of the two-sided confidence interval on the difference in means?

Your Answer:

Two types of plastics are suitable for use by an electronics component manufacturer. The breaking strength of these plastics is very important. From a random sample size of n₁=18, and n₂=16, we obtained that X₁-bar=151.2, S₁=1.4 and X₂-bar=152.3, S₂=1.65.

Calculate a 95% confidence interval on the difference in means, assuming that the population variances are equal.

With 95% confidence, what is the right-value of the two-sided confidence interval on the difference in means?

Your Answer:

Answer 1

A) At 95% confidence interval the critical values are

F() = F(0.025, 17, 15) = 2.813

F() = F(0.975, 17,15) = 0.367

The 95% confidence interval for is

<  <

= (1.4)^2/(1.65)^2 * (1/2.813) < <  (1.4)^2/(1.65)^2 * (1/0.367)

= 0.2559 < < 0.2642

So the right value of the 95% confidence interval is 0.2642

B) The pooled variance(s_p²) = ((n1 - 1)s1^2 + (n2 - 1)s2^2) /(n1 + n2 - 2) = (17 * (1.4)^2 + 15 * (1.65)^2)/(18 + 16 - 2) = 2.3174

Df = 18 + 16 - 2 = 32

At 95% confidence interval the critical value is t^* = 2.037

The 95% confidence interval for is

() +/- t^* * sqrt(s_p²/n1 + s_p²/n2)

= (151.2 - 152.3) +/- 2.037 * sqrt(2.3174/18 + 2.3174/16)

= -1.1 +/+ 1.0655

= -2.1655, -0.0345

The left value of the 95% confidence interval is -2. 1655.

The right value of the 95% confidence interval is -0.0345