The starting salaries of a random sample of three students who graduated from No

Question

The starting salaries of a random sample of three students who graduated from North Carolina State University last year with majors in the mathematical sciences are $47,000, $30,000, and $52,000. (a) Using software or a calculator, find the 95% confidence interval for the population mean. (b) Name two things you could do to get a narrower interval than the one in (a) O decrease the confidence level O increase the confidence interval odecrease the sample size O increase the sample size (c) Construct a 99% confidence interval. why is it wider than the 95% interval? O The margin of error is larger for a higher confidence level O The margin of error is smaller for a higher confidence level. O The t-value is smaller for a higher confidence level. (d) On what assumptions is the interval in (a) based? O The sample size is large. O The population distribution is uniform O The population distribution is approximately normal O The data is obtained by randomization.

Explanation / Answer

PART A.
TRADITIONAL METHOD
given that,
sample mean, x =43000
standard deviation, s =11532.5626
sample size, n =3
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 11532.5626/ sqrt ( 3) )
= 6658.328
II.
margin of error = t /2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 2 d.f is 4.303
margin of error = 4.303 * 6658.328
= 28650.786
III.
CI = x ± margin of error
confidence interval = [ 43000 ± 28650.786 ]
= [ 14349.214 , 71650.786 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =43000
standard deviation, s =11532.5626
sample size, n =3
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 2 d.f is 4.303
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 = [ 43000 ± t a/2 ( 11532.5626/ Sqrt ( 3) ]
= [ 43000-(4.303 * 6658.328) , 43000+(4.303 * 6658.328) ]
= [ 14349.214 , 71650.786 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ 14349.214 , 71650.786 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population mean

PART B.
decrease the confidence interval
increase the sample size

PART C.
At 0.01 LOS
confidence interval = [ 43000 ± t a/2 ( 11532.5626/ Sqrt ( 3) ]
= [ 43000-(9.925 * 6658.328) , 43000+(9.925 * 6658.328) ]
= [ -23083.907 , 109083.907 ]

The margin of error is wider for the confidence interval

PART D.
The population distribution is approximately normal

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