Question

The actual width of a 2×4 piece of lumber is approximately 1 3/4 in (1.75 in). but can vary considerably. Larry’s Lumber advertises consistent dimensions for better building, and claims all 2×4 sold have a mean width of 1 3/4 in. with a standard deviation of 0.02 in.

a. [5 marks] Assume the distribution of widths is approximately normal. Find a symmetric interval about the mean that contains almost all of the 2×4 widths. Show your work. (1.75

b. [5 marks] Suppose a random 2×4 has width 1.78 in. Is there any evidence to suggest the Larry’s Lumber claim is wrong? Justify your answer.

c. [5 marks] Suppose a random 2×4 has width 1.68 in. Is there any evidence to suggest the Larry’s Lumber claim is wrong? Justify your answer.

Answer 1

Answer and Explanation:

Let X denotes the actual width of a 2×4 piece of lumber in inches.

a. A symmetric interval about the mean that contains almost all of the 2×4 widths is calculated as:

Hence, the symmetric interval about the mean that contains almost all of the 2×4 widths is (1.69,1.81).

b.

Let the width of a random 2×4 piece of lumber be denoted by .

The null and the alternative hypothesis is denoted by respectively and is given by:

The test is a two tailed test.

The significance level is unknown so we take it as 5 percent.

The tabulated value of Z at percent level of significance is .

Under Null Hypothesis:

The value of associated test statistic for this test is given by:

Since, the calculated value of Z is less than tabulated value of Z. (-1.5>-1.96) and (1.5<1.96)

We conclude that Larry Lumber claim is correct.

c.

Let the width of a random 2×4 piece of lumber be denoted by .

The null and the alternative hypothesis is denoted by respectively and is given by:

The test is a two tailed test.

The significance level is unknown so we take it as 5 percent.

The tabulated value of Z at percent level of significance is .

Under Null Hypothesis:

The value of associated test statistic for this test is given by:

Since, the calculated value of Z is greater than tabulated value of Z. (3.5>1.96)

We conclude that Larry Lumber claim is wrong.