I need help with this problem!!! It’s confusing me so much and I need it by tomorrow!please tell me the answer and how you got i
t!
1 answer:
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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).
Answer:
$0.45x
Step-by-step explanation:
Fred earns dollars. Let Fred earns $x.
Chad earns 45% more than Fred.
We need to find the expression for Chad.
Chad amount = 45% of x
So, Chad earns $0.45x.