A random sample of the closing stock prices in dollars for a company in a recent

Question

A random sample of the closing stock prices in dollars for a company in a recent year is listed below. Assume that is $2.61. Construct the 90% and 99% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals 18.49 17.14 22.25 18.49 18.57 20.59 18.79 15.96 17.44 18.62 22.88 16.71 15.67 20.95 15.09 22.79 The 90% confidence interval is (soo (Round to two decimal places as needed.) The 99% confidence interval is (soo (Round to two decimal places as needed.) Which statement below interprets the results correctly? There is 90% confidence that the mean closing stock price is in the 90% confidence interval and 99% confidence that the mean closing stock price is in the 99% confidence interval The 90% confidence interval contains the mean closing stock price 90% of the time and the 99% confidence interval contains the mean closing stock price 99% of the time ( 90% of the mean closing stock prices are in the 90% confidence interval and 99% of the mean closing stock prices are in the 99% confidence interval The probability that the mean closing stock price is in the 90% confidence interval is about 90% and the probability that the mean closing stock price is in the 99% confidence interval is about 99% Which interval is wider? The 90% confidence interval The 99% confidence interval

Explanation / Answer

Q1.
AT 90% CI
TRADITIONAL METHOD
given that,
sample mean, x =18.7769
standard deviation, s =2.4922
sample size, n =16
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 2.4922/ sqrt ( 16) )
= 0.623
II.
margin of error = t /2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, = 0.1
from standard normal table, two tailed value of |t /2| with n-1 = 15 d.f is 1.753
margin of error = 1.753 * 0.623
= 1.092
III.
CI = x ± margin of error
confidence interval = [ 18.7769 ± 1.092 ]
= [ 17.685 , 19.869 ]
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DIRECT METHOD
given that,
sample mean, x =18.7769
standard deviation, s =2.4922
sample size, n =16
level of significance, = 0.1
from standard normal table, two tailed value of |t /2| with n-1 = 15 d.f is 1.753
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 18.7769 ± t a/2 ( 2.4922/ Sqrt ( 16) ]
= [ 18.7769-(1.753 * 0.623) , 18.7769+(1.753 * 0.623) ]
= [ 17.685 , 19.869 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 90% sure that the interval [ 17.685 , 19.869 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contains the true population mean

Q2.
AT 99% CI
confidence interval = [ 18.7769 ± t a/2 ( 2.4922/ Sqrt ( 16) ]
= [ 18.7769-(2.947 * 0.623) , 18.7769+(2.947 * 0.623) ]
= [ 16.941 , 20.613 ]

Q3.
The 90% confidence interval contains the mean closing stock price 90% of the time and the 99% confidence interval contains the mean closing stock price 99%
of the time.
Option B

Q4.
The 99% confidence interval is wider

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