You are the foreman of the Bar-S cattle ranch in Colorado. A neighboring ranch has calves for sale, and you are going to buy som
e calves to add to the Bar-S herd. How much should a healthy calf weigh? Let x be the age of the calf (in weeks), and let y be the weight of the calf (in kilograms).
x 3 4 11 16 26 36
y 38 51 74 100 150 200
Complete parts (a) through (e), given Σx = 96, Σy = 613, Σx^2 = 2374, Σy^2 = 82,021, Σxy = 13,832, and r = 0.998
(a) Draw a scatter diagram displaying the data
x 3 4 11 16 26 36
y 38 51 74 100 150 200
Complete parts (a) through (e), given Σx = 96, Σy = 613, Σx^2 = 2374, Σy^2 = 82,021, Σxy = 13,832, and r = 0.998
(a) Draw a scatter diagram displaying the data
1 answer:
Answer:
When you have two sets of data, one that represents an independent variable (X) and another that represents a dependent variable (Y) (it is a response of the first variable). A diagram called scatterplot can be generated to present the ordered pairs (X, Y) in the Cartesian plane to see how the two variables are related.
In this case the independent variable (X) is the age variable of the calf and the dependent variable (Y) is the weight of the calf.
The diagram can be seen in the attached file.
Step-by-step explanation:
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Answer:
a) In the step-by-step explanation.
b) The null and alternative hypothesis are:
c) P-value = 0.015
d) The conclusion, at a level of significance of 0.05, is that there is enough evidence to support the claim that that area residential property is available at a mean cost significantly higher than $125,000 per lot.
e) The claim of the chamber of commerce is false. We have statistical evidence against that claim and we can conclude that the mean prices are significantly higher than $125,000.
Step-by-step explanation:
This is a hypothesis test for the population mean.
The claim, expressed in the alternative hypothesis, is that area residential property is available at a mean cost significantly higher than $125,000 per lot.
Then, the null and alternative hypothesis are:
The significance level is 0.05.
The sample has a size n=32.
The sample mean is M=130000.
As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=12500.
The estimated standard error of the mean is computed using the formula:
Then, we can calculate the t-statistic as:
The degrees of freedom for this sample size are:
This test is a right-tailed test, with 31 degrees of freedom and t=2.26, so the P-value for this test is calculated as (using a t-table):
As the P-value (0.015) is smaller than the significance level (0.05), the effect is significant.
The null hypothesis is rejected.
There is enough evidence to support the claim that that area residential property is available at a mean cost significantly higher than $125,000 per lot.
The maximum number of adults is 4
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