Answer:
The 99% (two-sided) confidence interval for the true average echo duration μ is between 0 sec and 1.99 sec.
Step-by-step explanation:
We have the sample standard deviation, so we use the student 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 = 101 - 1 = 100
99% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 100 degrees of freedom(y-axis) and a confidence level of [tex]1 - \frac{1 - 0.99}{2} = 0.995. So we have T = 2.6259
The margin of error is:
M = T*s = 2.6259*0.45 = 1.18
In which s is the standard deviation of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 0.81 - 1.18. Answer in seconds cannot be negative, so we use 0 sec.
The upper end of the interval is the sample mean added to M. So it is 0.81 + 1.18 = 1.99 sec
The 99% (two-sided) confidence interval for the true average echo duration μ is between 0 sec and 1.99 sec.
I am not entirely sure if this is correct, so forgive me if I am wrong
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Answer: 266,667 bottles
Step-by-step explanation:
- 80/300 people preferred soft drink z. This as a decimal is 0.2666
This means that out of the million, 0.2666666666666667% need to be soft drink z
1,000,000x0.2666666666666667 =266,666.667
This can be rounded to 266,667 bottles of soft drink z
(I chose to include the whole decimal for the most accurate results, and save rounding for the last step)
Work is attached. Hope it helps!
Answer: the following points are in Quadrant 3
Answer:
need jo rin po ng sagot
Step-by-step explanation:
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