please response to number 1 and 2 individually. I will put a sample reply for you to follow. thank you so much

1. For Scenario 1, I will choose to compute a Z value. This is because we know the population standard deviation, which is 7.5.

Z = X/n = 31.829.67.5/30 = 2.27.5/5.477 = 2.21.37 = 1.6 (Sample 1)

Z = X/n = 31.629.67.5/50 = 27.5/7.07 = 21.06 = 1.9 (Sample 2)

None of these samples are concerning since they are within normal limits of the Z-curve.

For Scenario 2, I will choose to compute a T value. This is because we do not know the population standard deviation.

T = Xs/n = 27.629.66.8/30 = 26.8/5.48 = 21.24 = -1.6 (Sample 1)

T = Xs/n = 28.029.67.7/50 = 1.67.7/7.07 = 1.61.09 = -1.5 (Sample 2)

None of these samples are concerning since they are within normal limits of the Z-curve.

2. For scenario 1 I believe we would use a Z valuestatistic to compute our problem. This is because we have all of the values that we need such as the population, x-value, and a standard deviation.I plugged in the values(31.8-29.6)/(7.5/sqrt(30), and came up with an answer of 1.606 in sample 1. For sample 2 in scenario 1, my answer came out to be 1.885 when I plugged in the values using (31.6-29.6)/(7.5/sqrt(50). I don't believe any of these values are to be of concern since neither is greater than the Z-value.

Since scenario 2 does not have a standard deviation value for our problem, we would go with using a T value statistic.Using the T value, I plugged in the values for sample 1 (27.6-29.6)/(6.8/sqrt(30) and got -1.6109. And for sample 2 my answer is -1.4693 using (28-29.6)/(7.7/sqrt(50), however I still do not think it is a concern because both samples show that they are not extreme or go over the T-value.

sample reply for 1

Hi,

I love how you accurately computed the z-statistic and t-statistic. Also, I agree with how you determined when to use the two formulas and how you interpreted the computed values. Great work!

sample reply for 2

Hi

I also chose to compute the Z value for scenario 1 because the population's standard deviation was given along with the other values needed such as the x-value and standard deviation. As for scenario 2, I also chose to compute the T value and agree that the results are not a concern because both samples showed not extreme results. The negative results mean that samples were below the mean average.