Jeffrey and Frank walk at different speeds.Frank’s walking speed can be represented by the equation y=85x, where is the time in
minutes and y is the distance in meters. The distance Jeffrey walked over the time is shown on the graph below
1 answer:
Answer:
Option (C)
Option (C)
Option (A)
Step-by-step explanation:
Part A.
Frank's walking speed can be represented by the equation.
Equation representing the speed is,
y = 85x
Therefore, walking speed of Frank = 85 meters per minute.
Option (C) will be the answer.
Part B.
Distance walked over the time by Jeffrey is given by the graph attached.
Equation for the speed will be,
y =
y = 80x
Therefore, walking speed of Jeffrey = 80 meters per minute
Difference in the speed of Frank and Jeffrey = 85 - 80 = 5 meters per minute.
Jeffery walks 5 meters per minute slower than Frank.
Option (C) will be the answer.
Part (C)
Distance traveled by Frank in 16 minutes,
y = 85(16) = 1360 meters
Time taken to travel the same distance by Jeffrey will be,
1360 = 80x
x = meters per minute
Option (A) will be the answer.
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Answer:
If we compare the p value and a significance level assumed we see that
so we can conclude that we reject the null hypothesis, and the actual true mean is higher than 210250 at 5% of significance.
Step-by-step explanation:
Data given and notation
represent the sample mean
represent the standard deviation for the sample
sample size
represent the value that we want to test
represent the significance level for the hypothesis test.
t would represent the statistic (variable of interest)
represent the p value for the test (variable of interest)
State the null and alternative hypotheses to be tested
We need to conduct a hypothesis in order to determine if the mean is higher than 210250, the system of hypothesis would be:
Null hypothesis:
Alternative hypothesis:
Compute the test statistic
We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:
(1)
t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
We can replace in formula (1) the info given like this:
Now we need to find the degrees of freedom for the t distirbution given by:
Conclusion
Since is a one right tailed test the p value would be:
If we compare the p value and a significance level assumed we see that
so we can conclude that we reject the null hypothesis, and the actual true mean is higher than 210250 at 5% of significance.