Mark jogs 10 miles in 2 hours.

A new dental bondlng agent. When bonding teeth, orthodontists must maintain a dry field. A new bonding adhesive(called "Smartbon

Gemiola [76]

Answer:

(a) H₀: μ ≥ 5.70 vs. Hₐ:μ < 5.70

(b) The rejection region is (t₀.₀₁,₉ ≤ -2.821).

(c) The value of the test statistic is -4.33.

(d) The true mean breaking strength of the new bonding adhesive is less than 5.70 Mpa.

Step-by-step explanation:

A hypothesis test should be conducted to determine that the if the true mean breaking strength of the new bonding adhesive is less than 5.70 Mpa.

(a)

The hypothesis is:

H₀: The true mean breaking strength of the new bonding adhesive is not less than 5.70 Mpa, i.e. μ ≥ 5.70.

Hₐ: The true mean breaking strength of the new bonding adhesive is less than 5.70 Mpa, i.e. μ < 5.70.

(b)

The alternate hypothesis indicates that the hypothesis test is left-tailed.

The rejection region for the left tailed test will be towards the lower tail of the t-distribution curve.

The significance level of the test is: α = 0.01.

The critical value is:

$t_{\alpha ,(n-1)}=t_{0.01,(10-1)}=t_{0.01,9}$

Use the t-table for the critical value.

$t_{\alpha ,(n-1)}=t_{0.01,9}=-2.821$

Since rejection region is in the lower tail the critical value will be negative.

Thus, the rejection region is (t₀.₀₁,₉ ≤ -2.821).

(c)

The test statistic value is:

$t=\frac{\bar x-\mu}{s/\sqrt{n}}$

Given:

$\bar x=5.07\\s=0.46\\n=10\\\mu=5.70$

Compute the value of the t-statistic as follows:

$t=\frac{\bar x-\mu}{s/\sqrt{n}}=\frac{5.07-5.70}{0.46/\sqrt{10}} =-4.33$

The value of the test statistic is -4.33.

(d)

The value of the test is less than the critical value.

$t=-4.33$

This implies that the test statistic lies in the rejection region.

Hence the null hypothesis will be rejected at 1% significance level.

Conclusion:

As the null hypothesis is rejected it can be concluded that the true mean breaking strength of the new bonding adhesive is less than 5.70 Mpa.

(e)

The conditions required for the t-test for single mean to be valid is:

The data should be continuous.
The parent population should be normally distributed.
The sample should be randomly selected.

5 0

3 years ago

BRAINLIESTTTTTT ASAPPPP PLEASE ANSWER MY QUESTION

yawa3891 [41]

The red line is represented by y ≥ -3x + 3.

The blue line is represented by y < 3/2x - 6.

A) In this system of inequalities, there is one solid line and one dashed line.

The solid line is the inequality y ≥ -3x + 3 because it has the symbol greater than or equal to. The "equal to" part of the symbol is important because it means that any coordinate located on the solid line is a solution.

However, the inequality y < 3/2x - 6 is a dashed line because it does not have "equal to." This makes the inequality a dashed line which means that all coordinates on it will not be a solution to the inequality.

In the inequality y ≥ -3x + 3, the shading is above the line. This means that all coordinates within the line and shaded area above the line will be a solution to the inequality.

In the inequality y < 3/2x - 6, the shading is below the line. This means that all coordinates within the shared area of the line will be a solution to the inequality. However, since y is not less than or EQUAL TO, all coordinates within the dashed line is not a solution to the inequality.

The solution area, where a coordinate is a solution to
both inequalities, is where the shaded areas of both inequalities overlap.

B) The point (-6, 3) is a not solution to both of the inequalities because it is not within the shared areas of the two inequalities.

Substitute the coordinate (-6, 3) in the inequality y < 3/2x - 6.

y < 3/2x - 6

3 < 3/2(-6) - 6

3 < -9 - 6

3 < -15

This is not true because a negative number cannot be greater than a positive number. Therefore, this proves that the coordinate (-6, 3) is not a solution for the inequality y < 3/2x - 6.

Check if the coordinate fits for the other inequality.

y ≥ -3x + 3

3 ≥ -3(-6) + 3

3 ≥ 18 + 3

3 ≥ 21

This is also not true because 3 is not greater than or equal to 21. Therefore, this coordinate is not a solution to the inequality y ≥ -3x + 3.

The coordinate (-6, 3) is not a solution for the system of inequalities because it is not a solution for both inequalities.