Kaitlyn gave 1/2 of her candy bar to arianna. arianna gave 1/3 of the candy she got from kaitlyn to cameron. what fraction of a
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Answer:
A 90% confidence interval of the true mean is [$119.86, $123.34].
Step-by-step explanation:
We are given that an irate student complained that the cost of textbooks was too high. He randomly surveyed 36 other students and found that the mean amount of money spent on textbooks was $121.60.
Also, the standard deviation of the population was $6.36.
Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;
P.Q. = ~ N(0,1)
where, = sample mean amount of money spent on textbooks = $121.60
= population standard deviation = $6.36
n = sample of students = 36
= population mean
<em>Here for constructing a 90% confidence interval we have used One-sample z-test statistics as we know about population standard deviation.</em>
<em />
So, 95% confidence interval for the population mean, is ;
P(-1.645 < N(0,1) < 1.645) = 0.90 {As the critical value of z at 5% level
of significance are -1.645 & 1.645}
P(-1.645 < < 1.645) = 0.90
P( <
<
) = 0.90
P( <
<
) = 0.90
<u>90% confidence interval for</u> = [
,
]
= [ ,
]
= [$119.86, $123.34]
Therefore, a 90% confidence interval of the true mean is [$119.86, $123.34].
Answer: The lower bound of confidence interval would be 0.116.
Step-by-step explanation:
Since we have given that
p = 13.2%= 0.132
n = 1105
At 90% confidence,
z = 1.645
So, Margin of error would be
So, the lower bound of the confidence interval would be
Hence, the lower bound of confidence interval would be 0.116.