Question

In: Statistics and Probability

Lazurus Steel Corporation produces iron rods that are supposed to be 36 inches long. The machine...

Lazurus Steel Corporation produces iron rods that are supposed to be 36 inches long. The machine that makes these rods does not produce each rod exactly 36 inches long. The lengths of the rods are normally distributed and vary slightly. It is known that when the machine is working properly, the mean length of the rods is 36 inches. The standard deviation of the lengths of all rods produced on this machine is always equal to 0.035 inch. The quality control department at the company takes a sample of 20 such rods every week, calculates the mean length of these rods, and tests the null hypothesis, μ=36 inches, against the alternative hypothesis, μ≠36 inches. If the null hypothesis is rejected, the machine is stopped and adjusted. A recent sample of 20 rods produced a mean length of 36.015 inches.

Calculate the p-value for this test of hypothesis. Based on this p-value, will the quality control inspector decide to stop the machine and adjust it if he chooses the maximum probability of a Type I error to be 0.01?

Use the normal distribution table. Round your answer to four decimal places.

p-value =

The machine

does not need or needs

adjustment.

the tolerance is +/-2%

Solutions

Expert Solution

Solution:

Given: When the machine is working properly, the mean length of the rods is 36 inches.

thus

The standard deviation of the lengths of all rods produced on this machine is always equal to 0.035 inch.

that is:

the null hypothesis, H0: μ=36 inches, against the alternative hypothesis, H1: μ≠36 inches.

If the null hypothesis is rejected, the machine is stopped and adjusted.

Sample size = n = 20

Sample mean =

We have to calculate the p-value for this test of hypothesis.

Level of significance =

Find z test statistic:

Since H1 is ≠ type , this is two tailed test, p-value is given by:

For two tailed test , p-value is:

p-value = 2* P(Z > z test statistic) if z is positive

p-value = 2* P(Z < z test statistic) if z is negative

thus

p-value = 2* P(Z > z test statistic)

p-value = 2* P(Z > 1.92)

p-value = 2* [ 1 - P(Z < 1.92) ]

Look in z table for z = 1.9 and 0.02 and find corresponding area.

P( Z < 1.92) = 0.9726

thus

p-value = 2* [ 1 - P(Z < 1.92) ]

p-value = 2* [ 1 - 0.9726 ]

p-value = 2* 0.0274

p-value = 0.0548

Since p-value = 0.0548 > 0.01 level of significance, we fail to reject H0.

Thus

The machine does not need  adjustment


Related Solutions

Lazurus Steel Corporation produces iron rods that are supposed to be 36 inches long. The machine...
Lazurus Steel Corporation produces iron rods that are supposed to be 36 inches long. The machine that makes these rods does not produce each rod exactly 36 inches long. The lengths of the rods are normally distributed and vary slightly. It is known that when the machine is working properly, the mean length of the rods is 36 inches. The standard deviation of the lengths of all rods produced on this machine is always equal to 0.035 inch. The quality...
A steel factory produces iron rods that are supposed to be 36 inches long. The machine...
A steel factory produces iron rods that are supposed to be 36 inches long. The machine that makes these rods does not produce each rod exactly 36 inches long. The lengths of these rods vary slightly. It is known that when the machine is working properly, the mean length of the rods is 36 inches. According to design, the standard deviation of the lengths of all rods produced on this machine is always equal to .05 inches. The quality control...
Lazurus Steel Corporation produces iron rods that are supposed to be 33 inches long. The machine...
Lazurus Steel Corporation produces iron rods that are supposed to be 33 inches long. The machine that makes these rods does not produce each rod exactly 33 inches long. The lengths of the rods vary slightly. It is known that when the machine is working properly, the mean length of the rods made on this machine is 33 inches. The standard deviation of the lengths of all rods produced on this machine is always equal to 0.3 inch. The quality...
A rods manufacturer makes rods with a length that is supposed to be 19 inches. A...
A rods manufacturer makes rods with a length that is supposed to be 19 inches. A quality control technician sampled 21 rods and found that the sample mean length was 19.03 inches and the sample standard deviation was 0.11 inches. The technician claims that the mean rod length is more than 19 inches. What type of hypothesis test should be performed? What is the test statistic? What is the number of degrees of freedom? Does sufficient evidence exist at the...
A. A company produces steel rods. The lengths of the steel rods are normally distributed with...
A. A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 162.4-cm and a standard deviation of 0.6-cm. For shipment, 16 steel rods are bundled together. Find the probability that the average length of a randomly selected bundle of steel rods is less than 162.6-cm. P(¯xx¯ < 162.6-cm) = B. A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 246.8-cm and a standard...
A company produces steel rods. The lengths of the steel rods are normally distributed with a...
A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 167.1-cm and a standard deviation of 0.6-cm. For shipment, 6 steel rods are bundled together. Round all answers to four decimal places if necessary. What is the distribution of XX? XX ~ N( , ) What is the distribution of ¯xx¯? ¯xx¯ ~ N( , ) For a single randomly selected steel rod, find the probability that the length is between 166.9-cm...
A company produces steel rods. The lengths of the steel rods are normally distributed with a...
A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 195.2-cm and a standard deviation of 0.8-cm. For shipment, 22 steel rods are bundled together. Find the probability that the average length of a randomly selected bundle of steel rods is less than 195.1-cm. P(M < 195.1-cm) = Enter your answer as a number accurate to 4 decimal places. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are...
A company produces steel rods. The lengths of the steel rods are normally distributed with a...
A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 143.6-cm and a standard deviation of 0.8-cm. For shipment, 41 steel rods are bundled together. Round all answers to four decimal places if necessary. What is the distribution of X ? X ~ N(,) What is the distribution of ¯x ? ¯x ~ N(,) For a single randomly selected steel rod, find the probability that the length is between 143.4-cm and 143.5-cm....
A company produces steel rods. The lengths of the steel rods are normally distributed with a...
A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 222-cm and a standard deviation of 1.5-cm. For shipment, 9 steel rods are bundled together. Find the probability that the average length of a randomly selected bundle of steel rods is less than 222.1-cm. P(M < 222.1-cm) = Enter your answer as a number accurate to 4 decimal places. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are...
A company produces steel rods. The lengths of the steel rods are normally distributed with a...
A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 124.7-cm and a standard deviation of 1.6-cm. For shipment, 17 steel rods are bundled together. Find P83, which is the average length separating the smallest 83% bundles from the largest 17% bundles. P83 = -cm Please provide a step-by-step!
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT