The following graph models bristols heart rate, in beats per minute(bpm), over time, in minutes during the last 10 minutes of yo
ga class. Bristols heart rate is 70.
Is the graph that models Bristol's heart rate a parabola?
Is the graph that models Bristol's heart rate a parabola?
2 answers:
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Answer:
If we compare the p value and the significance level given we see that
so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the average tensile strength is different from 5lbs/mm at 10% of signficance
Step-by-step explanation:
Data given and notation
represent the sample mean
represent the population 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.
We need to conduct a hypothesis in order to check if the true mean is different from 5, the system of hypothesis would be:
Null hypothesis:
Alternative hypothesis:
If we analyze the size for the sample is > 30 but and we know the population deviation so is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:
(1)
z-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".
Calculate the statistic
We can replace in formula (1) the info given like this:
P-value
Since is a two sided test the p value would be:
Conclusion
If we compare the p value and the significance level given we see that
so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the average tensile strength is different from 5lbs/mm at 10% of signficance
There were 340,000 cattle placed on feed
How many of the 340,000 cattle placed on feed were between 700 and 799 pounds?
Given the fraction of total cattle for 700 - 799 pounds is 2/5
Let x be the number of cattle between 700 - 799 pounds
We make a proportion using the fraction
Cross multiply it and solve for x
340000* 2 = 5x
680000 = 5x
Divide by 5 on both sides
So x= 136,000
There were 136,000 cattle between 700 and 799 pounds