a computer sales associate makes $74 each day that he works and makes approximately $20 in commission for each computer that he
1 answer:
You might be interested in
Answer:
The proportions differ from those reported in the survey.
Step-by-step explanation:
The Chi-square goodness of fit test would be used to determine whether the proportions differ from those reported in the survey.
The hypothesis for the test can be defined as follows:
<em>H</em>₀: The proportions does not differ from those reported in the survey.
<em>Hₐ</em>: The proportions differ from those reported in the survey.
Assume that the significance level of the test is, α = 0.01.
The Chi-square test statistic is given by:
Consider the Excel sheet provided.
The Chi-square test statistic value is 191.32.
The <em>p</em>-value of the test is:
The <em>p</em>-value of the test is very small. The null hypothesis will be rejected at 1% level of significance.
Thus, concluding that the proportions differ from those reported in the survey.
Answer:
Since the significance level is 0.05 we see that so we have enough evidence to reject the null hypothesis. And the best conclusion for this case would be:
b. at least some, but not all, of the gestation periods across all four species are the same
Because is only to identify if AT LEAST one mean is different, NOT to conclude that the all the means are different.
Step-by-step explanation:
Previous concepts
Analysis of variance (ANOVA) "is used to analyze the differences among group means in a sample".
The sum of squares "is the sum of the square of variation, where variation is defined as the spread between each individual value and the grand mean"
Solution to the problem
The hypothesis for this case are:
Null hypothesis:
Alternative hypothesis: Not all the means are equal
In order to find the mean square between treatments (MSTR), we need to find first the sum of squares and the degrees of freedom.
If we assume that we have groups and on each group from
we have
individuals on each group we can define the following formulas of variation:
And we have this property
And in order to test this hypothesis we need to ue an F statistic and for this case the p value calculated is
Since the significance level is 0.05 we see that so we have enough evidence to reject the null hypothesis. And the best conclusion for this case would be:
b. at least some, but not all, of the gestation periods across all four species are the same
Because is only to identify if AT LEAST one mean is different NOT to conclude that the all the means are different.
Answer:
(A) The probability that a randomly selected adult is either overweight or obese is 0.688.
(B) The probability that a randomly selected adult is neither overweight nor obese is 0.312.
(C) The events "overweight" and "obese" exhaustive.
(D) The events "overweight" and "obese" mutually exclusive.
Step-by-step explanation:
Denote the events as follows:
<em>X</em> = a person is overweight
<em>Y</em> = a person is obese.
The information provided is:
A person is overweight if they have BMI 25 or more but below 30.
A person is obese if they have BMI 30 or more.
P (X) = 0.331
P (Y) = 0.357
(A)
The events of a person being overweight or obese cannot occur together.
Since if a person is overweight they have (25 ≤ BMI < 30) and if they are obese they have BMI ≥ 30.
So, P (X ∩ Y) = 0.
Compute the probability that a randomly selected adult is either overweight or obese as follows:
Thus, the probability that a randomly selected adult is either overweight or obese is 0.688.
(B)
Commute the probability that a randomly selected adult is neither overweight nor obese as follows:
Thus, the probability that a randomly selected adult is neither overweight nor obese is 0.312.
(C)
If two events cannot occur together, but they form a sample space when combined are known as exhaustive events.
For example, flip of coin. On a flip of a coin, the flip turns as either Heads or Tails but never both. But together the event of getting a Heads and Tails form a sample space of a single flip of a coin.
In this case also, together the event of a person being overweight or obese forms a sample space of people who are heavier in general.
Thus, the events "overweight" and "obese" exhaustive.
(D)
Mutually exclusive events are those events that cannot occur at the same time.
The events of a person being overweight and obese are mutually exclusive.
Answer:
equation of a line:
y = mx+c
1) find the gradient, m
2) find y-intercept, c using coordinate (1,-4)
y = mx + c
-4 = 0(1) + c
c = -4
the equation of line:
y = mx+c
y = 0(x) + c
y = c
y = -4