1.A manufacturer of sports equipment has developed a new synthetic fishing line

Question

1.A manufacturer of sports equipment has developed a new synthetic fishing line that he claims to have a mean breaking strength of 8.1 kg, with standard deviation of 0.4. A random sample of 20 lines is tested and found to have a mean 7.5 kg. Is there sufficient evidence to reject the manufacturer claim at a = 0.01?

2.A manufacturer of sports equipment has developed a new synthetic fishing line that he claims to have a mean breaking strength of 8.1 kg. A random sample of 20 lines is tested and found to have a mean 7.5 kgwith standard deviation of 0.4. Is there sufficient evidence to reject the manufacturer claim at a = 0.01?

3.A commonly prescribed drug on the market for relieving nervous tension is believed to be 60% effective. Experimental results with a new drug administered to a random sample of 100 adults who suffer nervous tension showed that 68 received relief. At a 0.05 significance level (a = 0.05), Is there sufficient evidence to conclude that the new drug is superior to the one commonly used?

4.Five 2 cm thick experimental pieces of an insulation material were tested at random. Resulted compression (y) was measured recorded for the given amount of measured pressure. The data are listed below:

Pressure

Compression

1

1

2

2

3

2

4

3

5

5

Find the followings:

a.Fit a simple linear model relating compression y to the applied pressure x.

b.For the pressure- compression model, estimate the error variance.

c.A measure of the strength of the linear relationship between x, and y in the sample.

Pressure

Compression

1

1

2

2

3

2

4

3

5

5

Explanation / Answer

1. Null Hypothesis, H0: µ = 8.1

Alternate Hypothesis,Ha: µ 8.1

Sample mean = 7.5

Sample variance estimate = 0.4 / sqrt(20) = 0.0894

The z-value corresponding to sample mean = (7.5-8.1)/0.0894 = -6.71

p-value = 9.64 x 10-12 < 0.015/2 i.e., 0.005

So we can reject the Null Hypothesis and there is sufficient evidence to reject the manufacturer claim.

2.

Null Hypothesis, H0: µ = 8.1

Alternate Hypothesis,Ha: µ 8.1

Sample mean = 7.5

Sample variance estimate = 0.4

The z-value corresponding to sample mean = (7.5-8.1)/0.4 = -1.5

p-value = 0.0668 > 0.01/2 i.e., 0.005

So we can not reject the Null Hypothesis and there isn't sufficient evidence to reject the manufacturer claim.

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