Answer:
B and C
Step-by-step explanation:
Both of these equationa can be used to find the length of each pencil
According to a daylight calculator for Fairbanks, AK, sun rises at 07:40 and sun sets at 20:12, giving 12 hours and 32 minutes of daylight.
Note that it is not a mistake that daylight is not exactly 12 hours on that day due to refraction of the sun's rays at sunrise and sunset.
At Fairbanks, near equality to 12 hours occurs on March 16 with sunrise at 07:58 and sunset at 19:57.
Answer:
= 1.56
Step-by-step explanation:
Hello!
You have two independent samples of fertilizers from distributor A and B, and need to test if the average content of nitrogen from fertilizer distributed by A is greater than the average content of nitrogen from fertilizer distributed by B.
Sample 1
X₁: Content of nitrogen of a fertilizer batch distributed by A
n₁= 4 batches
Sample mean X₁[bar]= 23pound/batch
σ₁= 4 poundes/batch
Sample 2
X₂: Content of nitrogen of a fertilizer batch distributed by B
n₂= 4 batches
Sample mean X₂[bar]= 18pound/batch
σ₂= 5 pounds/batch
Both variables have a normal distribution. Now since you have the information on the variables distribution and the values of the population standard deviations, you could use a pooled Z-test.
Your hypothesis are:
H₀: μ₁ ≤ μ₂
H₁: μ₁ > μ₂
The statistic to use is:
Z= <u> X₁[bar] - X₂[bar] - (μ₁ - μ₂) </u>~N(0;1)
√(δ²₁/n₁ + δ²₂/n₂)
=<u> </u>(<u>23 - 18) - 0 </u> = 1.56
√(16/4 + 25/4)
I hope this helps!
Answer:
a
Step-by-step explanation:
Answer:
Step-by-step explanation:
From the given information, we can compute the table showing the summarized statistics of the two alloys A & B:
Alloy A Alloy B
Sample mean

Equal standard deviation

Sample size

Mean of the sampling distribution is :

Standard deviation of sampling distribution:

Hypothesis testing.
Null hypothesis: 
Alternative hypothesis: 
The required probability is:
![P(\overline X_A - \overline X_B>4|\mu_A - \mu_B) = P\Big (\dfrac{(\overline X_A - \overline X_B)-\mu_{X_A-X_B}}{\sigma_{\overline x_A -\overline x_B}} > \dfrac{4 - \mu_{X_A-\overline X_B}}{\sigma _{\overline x_A - \overline X_B}} \Big) \\ \\ = P \Big( z > \dfrac{4-0}{1.2909}\Big) \\ \\ = P(z \ge 3.10)\\ \\ = 1 - P(z < 3.10) \\ \\ \text{Using EXCEL Function:} \\ \\ = 1 - [NORMDIST(3.10)] \\ \\ = 1- 0.999032 \\ \\ 0.000968 \\ \\ \simeq 0.0010](https://tex.z-dn.net/?f=P%28%5Coverline%20X_A%20-%20%5Coverline%20X_B%3E4%7C%5Cmu_A%20-%20%5Cmu_B%29%20%3D%20P%5CBig%20%28%5Cdfrac%7B%28%5Coverline%20X_A%20-%20%5Coverline%20X_B%29-%5Cmu_%7BX_A-X_B%7D%7D%7B%5Csigma_%7B%5Coverline%20x_A%20-%5Coverline%20x_B%7D%7D%20%3E%20%5Cdfrac%7B4%20-%20%5Cmu_%7BX_A-%5Coverline%20X_B%7D%7D%7B%5Csigma%20_%7B%5Coverline%20x_A%20-%20%5Coverline%20X_B%7D%7D%20%20%20%5CBig%29%20%5C%5C%20%5C%5C%20%3D%20P%20%5CBig%28%20z%20%3E%20%5Cdfrac%7B4-0%7D%7B1.2909%7D%5CBig%29%20%5C%5C%20%5C%5C%20%3D%20P%28z%20%5Cge%203.10%29%5C%5C%20%5C%5C%20%3D%201%20-%20P%28z%20%3C%203.10%29%20%5C%5C%20%5C%5C%20%5Ctext%7BUsing%20EXCEL%20Function%3A%7D%20%5C%5C%20%5C%5C%20%20%3D%201%20-%20%5BNORMDIST%283.10%29%5D%20%20%5C%5C%20%5C%5C%20%3D%201-%200.999032%20%5C%5C%20%5C%5C%200.000968%20%5C%5C%20%5C%5C%20%5Csimeq%20%200.0010)
This implies that a minimal chance of probability shows that the difference of 4 is not likely, provided that the two population means are the same.
b)
Since the P-value is very small which is lower than any level of significance.
Then, we reject
and conclude that there is enough evidence to fully support alloy A.