Suppose that 500 randomly selected alumni of the University of Okoboji were aske

Question

Suppose that 500 randomly selected alumni of the University of Okoboji were asked to rate the university's academic advising services on a 1 to 10 scale. The sample mean x was found to be 8.6. Assume that the population standard deviation is known to be sigma = 2.2. Bitlost computes the 95% confidence interval for the average satisfaction score as 8.6 plusminus 1.96(2.2). What is her mistake? After correcting her mistake in part (a), she states, "I am 95% confident that the sample mean falls between 8.4 and 8.8." What is wrong with this statement? She quickly realizes her mistake in part (b) and instead states, "The probability that the true mean is between 8.4 and 8.8 is 0.95." What misinterpretation is she making now Finally, in her defense for using the Normal distribution to determine the confidence interval she says, "Because the sample size is quite large, the population of alumni ratings will be approximately Normal." Explain to her misunderstanding, and correct this statement.

Explanation / Answer

a.
Confidence Interval
CI = x ± Z a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=8.6
Standard deviation( sd )=2.2
Sample Size(n)=500
Confidence Interval = [ 8.6 ± Z a/2 ( 2.2/ Sqrt ( 500) ) ]
= [ 8.6 ± 1.96 * ( 2.2/ Sqrt ( 500) ) ]

in context it is only multplied with 2.2

b.
We are 95% sure that the interval contains the true population mean it is n't sample

c.
the true mean is alwasys lies in between

d.
Yes it tends to normal as n>=30

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