The correlation between a car’s engine size and its fuel economy (in mpg) is r = - 0.774.
1 answer:
Answer: 59.91%.
Step-by-step explanation:
- We know that the fraction of the variability in data values accounted by a model is given by
, where r is the coefficient of correlation.
We are given , that the correlation between a car’s engine size and its fuel economy (in mpg) is r = - 0.774.
Then, the fraction of the variability in fuel economy is accounted for by the engine size would be
[Multiply 100 to convert a decimal into percent]
Hence, the fraction of the variability in fuel economy is accounted for by the engine size is 59.91%.
None of the options are correct.
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Answer:
The margin of error of u is of 3.8.
The 99% confidence interval for the population mean u is between 27.4 minutes and 35 minutes.
Step-by-step explanation:
We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 28 - 1 = 27
99% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 27 degrees of freedom(y-axis) and a confidence level of . So we have T = 2.7707
The margin of error is:
In which s is the standard deviation of the sample and n is the size of the sample.
The margin of error of u is of 3.8.
The lower end of the interval is the sample mean subtracted by M. So it is 31.2 - 3.8 = 27.4 minutes
The upper end of the interval is the sample mean added to M. So it is 31.2 + 3.8 = 35 minutes
The 99% confidence interval for the population mean u is between 27.4 minutes and 35 minutes.