In the table below, x represents miles traveled and y represents the cost to travel by train.
1 answer:
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Answer:
The 90% confidence interval for the mean test score is between 77.29 and 85.71.
Step-by-step explanation:
We have the standard deviation for the sample, so we use the t-distribution 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 = 25 - 1 = 24
90% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 24 degrees of freedom(y-axis) and a confidence level of . So we have T = 2.064
The margin of error is:
In which s is the standard deviation of the sample and n is the size of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 81.5 - 4.21 = 77.29
The upper end of the interval is the sample mean added to M. So it is 81.5 + 4.21 = 85.71.
The 90% confidence interval for the mean test score is between 77.29 and 85.71.
The value of is 31
Explanation:
The given expression is
We need to determine the value of
The value of can be determined by substituting
in the expression
Thus, we have,
Adding the terms, we have,
Thus, the value is 31.
Therefore, the simplified value of the expression by substituting in the expression
is 31.