The strength of a certain type of rubber is tested by subjecting pieces of the rubber to an abrasion test. for the rubber to be acceptable, the mean weight loss μ must be less than 3.5 mg. a large number of pieces of rubber that were cured in a certain way were subject to the abrasion test. a 95% upper confidence bound for the mean weight loss was computed from these data to be 3.45 mg. someone suggests using these data to test h0 : μ ≥ 3.5 versus h1 : μ < 3.5. it is discovered that the mean of the sample used to compute the confidence bound is x⎯⎯⎯ = 3.40. is it possible to determine whether p < 0.01? explain. round the test statistic to two decimal places and the answer to four decimal places.

Step-by-step explanation:

Hello!

You have the hypothesis that the average weight loss for rubber after an abrasion test is less than 3.5 mg. To test this a large sample of pieces of rubber were sampled and subjected to the abrasion test.

With the given information you must test whether the researcher's hypothesis is sustained or not.

The study variable is,

X: Weight loss of rubber cured in a certain way after being subjected to the abrasion test. (mg)

There is no information about the variable distribution, but since it is said that the sample is a "large number" I'll take it as if it is bigger than 30 and apply the Central Limit Theorem to use the approximation of the sample mean to normal. This way I can use the Z-statistic for the test.

Symbolically the statistic hypothesis is:

H₀: μ ≥ 3.5

H₁: μ < 3.5

α: 0.05 (since is not listed, I'll choose one of the most common signification levels)

You have a one-tailed critical region, this means the p-value will also be one-tailed to the left of the distribution (i.e. →-∞)

The formula of the statistic is:

Z= X[bar] - μ ≈ N(0;1)

        δ/√n

To calculate the statistic you have to use the information given.

The sample mean X[bar]= 3.4 mg

Upper bond of 95% CI= 3.45 mg

The basic structure of a CI for the mean is

"estimator" ± "margin of error"

Upper bound is "estimator" + "margin of error"

Using the formula:

Ub= X[bar] + d ⇒ 3.45= 3.4 + d

⇒ d= 3.45 - 3.4 = 0.05

Where d is the margin of error

d= * (δ/√n)

d= * (δ/√n)

d/= (δ/√n)

(δ/√n)= 0.05/ 1.96 = 0.0255

(δ/√n) is the denominator in the formula, corresponds to the standard deviation of the distribution.

Now you have all values and can calculate the statistic under the null hypothesis:

Z= 3.4 - 3.5 = -3.92

       0.0255

And the p-value:

P(Z ≤ -3.92) = 0.000044 ⇒ My Z- table goes up to P(Z ≤ -3.00) = 0.001, so using strictly the table I can say that the probability is less than 0.001.

To calculate the exact probability I've used a statistic program.

p-value < 0.001

I hope it helps!

true

explanation:

not enough information

step-by-step explanation:

we cannot know this, because the other students could have handed out 10 flyers, or they could have handed out 100. one way, the mean would go up, and another way, the mean would go down.