A teacher believes that the third homework assignment is a key predictor in how well students will do on the midterm. Let x repr
1 answer:
Answer:
Hello your question has a disjointed equation attached to it and it is also incomplete attached below is the correct and complete question
A teacher believes that the third homework assignment is a key predictor in how well students will do on the midterm. Let x represent the third homework score and y the midterm exam score. A random sample of last terms students were selected and their grades are shown below
answer :
y-intercept = 25.9047
slope = 2.8420
Step-by-step explanation:
Determine the slope for the regression equation and y intercept
The regression equation ( gotten using excel ; attached below is the excel sample on how the equation was gotten )
y = 25.9047 + 2.8420x
from the equation gotten above
y-intercept = 25.9047
slope = 2.8420
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Answer:
The 95% confidence interval for the population's mean length of vacation, in days, is (4.24, 4.76).
The 92% confidence interval for the population's mean length of vacation, in days, is (4.27, 4.73).
Step-by-step explanation:
We have the standard deviations 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 = 85 - 1 = 84
95% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 84 degrees of freedom(y-axis) and a confidence level of . So we have T = 1.989.
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 4.5 - 0.26 = 4.24 days
The upper end of the interval is the sample mean added to M. So it is 4.5 + 0.26 = 4.76 days
The 95% confidence interval for the population's mean length of vacation, in days, is (4.24, 4.76).
92% confidence interval:
Following the sample logic, the critical value is 1.772. So
The lower end of the interval is the sample mean subtracted by M. So it is 4.5 - 0.23 = 4.27 days
The upper end of the interval is the sample mean added to M. So it is 4.5 + 0.23 = 4.73 days
The 92% confidence interval for the population's mean length of vacation, in days, is (4.27, 4.73).