Using the t-distribution, it is found that the 99% confidence interval for the difference is (0.058, 3.462).
We are given the <u>standard deviation for the samples</u>, hence, the t-distribution is used to solve this question.
The standard errors are given by:
For the distribution of differences, the <u>mean and the standard error</u> are given by:
The interval is given by:
The critical value for a <u>two-tailed 99% confidence interval</u> with 18 + 18 - 2 = <u>34 df</u> is t = 2.7284.
Then:
The 99% confidence interval for the difference is (0.058, 3.462).
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Answer:
[6 , 18]
Step-by-step explanation:
Answer:
58 per month
Step-by-step explanation:
<u>divide by 3</u>
<u>174 cars</u> = <u>58 cars</u>
3 months 1 month
<u> </u><u>divide by 3</u>
Answer: Double check some of them but other than that you have a few right-
Step-by-step explanation:
The probability that ticket was a student ticket is 42 / 132
Recall :
Probability = Required outcome / Total possible outcomes
To obtain the probability that the ticket sold for space force is a student ticket :
Total possible outcomes = number of ticket sold for space force = 132 tickets
Required outcome = number of student tickets sold for space force = 42
Hence,
P(space force | student ticket) = 42 / 132
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