Answer:
<em>H</em>₀: <em>μ</em> = 4 vs. <em>H
ₐ</em>: <em>μ </em>> 4
Step-by-step explanation:
A null hypothesis is a sort of hypothesis used in statistics that intends that no statistical significance exists in a set of given observations.
It is a hypothesis of no difference.
It is typically the hypothesis a scientist or experimenter will attempt to refute or discard. It is denoted by H₀.
Whereas, the alternate hypothesis is the contradicting statement to the null hypothesis.
The alternate hypothesis describes direction of the hypothesis test, i.e. if the test is left tailed, right tailed or two tailed.
It is also known as the research hypothesis and is denoted by H
ₐ.
In this case we need to test whether the amount is paid after the grace period, on average, more than 4 times in 2018.
The hypothesis can be defined as follows:
<em>H</em>₀: <em>μ</em> = 4 vs. <em>H
ₐ</em>: <em>μ </em>> 4
Answer: Independent event
Step-by-step explanation:
When two events are independent of each other, this means that the probability that one event will occurs does not in any way affects the probability of the occurrence of the other event.
For example, card is picked at random from a deck of cards, and after putting it back,thne another card is picked at random. The probability of picking the cards is not affected by each other since it is put back.
2.80*.15=.42
2.80-.42= 2.38
the answer is $2.38 (this was the price before)
Since he is using 3x more flower than the original, you would multiply the 2 tsp by 3 as well
6tsp is the answer
Answer:
a) 
b) 
c) 
With a frequency of 4
d) 
<u>e)</u>
And we can find the limits without any outliers using two deviations from the mean and we got:
And for this case we have two values above the upper limit so then we can conclude that 1500 and 3000 are potential outliers for this case
Step-by-step explanation:
We have the following data set given:
49 70 70 70 75 75 85 95 100 125 150 150 175 184 225 225 275 350 400 450 450 450 450 1500 3000
Part a
The mean can be calculated with this formula:
Replacing we got:
Part b
Since the sample size is n =25 we can calculate the median from the dataset ordered on increasing way. And for this case the median would be the value in the 13th position and we got:
Part c
The mode is the most repeated value in the sample and for this case is:
With a frequency of 4
Part d
The midrange for this case is defined as:
Part e
For this case we can calculate the deviation given by:
And replacing we got:
And we can find the limits without any outliers using two deviations from the mean and we got:
And for this case we have two values above the upper limit so then we can conclude that 1500 and 3000 are potential outliers for this case
