,Xn be a random sample from a normal distribution N(, 2). (a) Find a 90% confide
ID: 3051920 • Letter: #
Question
,Xn be a random sample from a normal distribution N(, 2). (a) Find a 90% confidence interval for based on the data for which X 19.3 and (b) By using the information in (a), find a one-sided lower 90% confidence interval (c) Suppose now that 2 is unknown. Find a 90% confidence interval for if X (d) If 2-9, what sample size is needed to achieve a 90% confidence interval for 5. Let Xi, n = 16 if it is known that 2-9. for . Also, find a one-sided upper 90% confidence interval for . 193, S2-1024, and n = 16. of length 2?Explanation / Answer
Given that,
Let X1,X2,....,Xn be a random sample from a normal distribution.
a) Now we have to find 90% confidence interval for population mean 9mu).
Given that,
Xbar = 19.3
n = 16
sigma2 = 9
sigma = sqrt(9) = 3
C-level = 90% = 0.90
Here population variance is known so we use one sample z-confidence interval.
We can find confidence interval in TI-83 calculator.
steps :
STAT --> TESTS --> 7:Z-interval --> ENTER --> Highlight on STATs --> ENTER --> Input all the values --> Calculate --> ENTER
90% confidence interval for mu is (18.066, 20.534).
c) Here sigma2 is unknown so we use one sample t-interval.
Given that,
Xbar = 19.3
S2 = 10.24
n = 16
S = sqrt(10.24) = 3.2
C = 90%
We can find one sample t-interval in TI-83 calculator.
steps :
STAT --> TESTS --> TInterval --> ENTER --> Highlight on Stats --> ENTER --> Input all values --> Calculate --> ENTER
90% confidence interval for population mean is (17.898, 20.702).
d) Now here we have to find sample size for :
sigma2 = 9
c = 90% = 0.9
length of confidence interval = 2
Now first we have to find margin of error (E).
E = 1/2 * length of confidence interval
E = 1/2 * 2 = 1
The formula for sample size is,
n = [ (Zc * sigma) / E ]2
where Zc is critical value for normal distribution.
Zc we can find in excel.
syntax :
=NORMSINV(probability)
where probability = 1 - a/2
where a = 1 - C
Zc = 1.645
n = [ (1.645 * 3) / 1]2 = 24.35
which is approximately equal to 24.
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