What is the probability that a point randomly chosen in the rectangle is in the triangle? A StartFraction 1 over 8 EndFraction B
2 answers:
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Answer:
(a) 80% confidence interval for the population mean is [109.24 , 116.76].
(b) 80% confidence interval for the population mean is [109.86 , 116.14].
(c) 98% confidence interval for the population mean is [105.56 , 120.44].
(d) No, we could not have computed the confidence intervals in parts (a)-(c) if the population had not been normally distributed.
Step-by-step explanation:
We are given that a simple random sample of size n is drawn from a population that is normally distributed.
The sample mean is found to be 113 and the sample standard deviation is found to be 10.
(a) The sample size given is n = 13.
Firstly, the pivotal quantity for 80% confidence interval for the population mean is given by;
P.Q. = ~
where, = sample mean = 113
s = sample standard deviation = 10
n = sample size = 13
= population mean
<em>Here for constructing 80% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.</em>
<u>So, 80% confidence interval for the population mean, </u><u> is ;</u>
P(-1.356 < < 1.356) = 0.80 {As the critical value of t at 12 degree
of freedom are -1.356 & 1.356 with P = 10%}
P(-1.356 < < 1.356) = 0.80
P( <
<
) = 0.80
P( <
<
) = 0.80
<u>80% confidence interval for </u> = [
,
]
= [ ,
]
= [109.24 , 116.76]
Therefore, 80% confidence interval for the population mean is [109.24 , 116.76].
(b) Now, the sample size has been changed to 18, i.e; n = 18.
So, the critical values of t at 17 degree of freedom would now be -1.333 & 1.333 with P = 10%.
<u>80% confidence interval for </u> = [
,
]
= [ ,
]
= [109.86 , 116.14]
Therefore, 80% confidence interval for the population mean is [109.86 , 116.14].
(c) Now, we have to construct 98% confidence interval with sample size, n = 13.
So, the critical values of t at 12 degree of freedom would now be -2.681 & 2.681 with P = 1%.
<u>98% confidence interval for </u> = [
,
]
= [ ,
]
= [105.56 , 120.44]
Therefore, 98% confidence interval for the population mean is [105.56 , 120.44].
(d) No, we could not have computed the confidence intervals in parts (a)-(c) if the population had not been normally distributed because t test statistics is used only when the data follows normal distribution.