Asked Oct 15, 2019

A random sample of size 30 from a normal population yields  = 39  and s = 4.9.  The lower bound of a 95 percent confidence interval is 

(Round off upto 2 decimal places). 

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Step 1


Hey there! Thank you for the question. In the first line, it is written “A random sample of size 30 from a normal population yields = 39”, which does not make sense. Logically, it can be the sample mean. Thus, we have assumed that the actual information is, = 39, and solved the problem.

Step 2


The 100 (1 – α) % confidence interval for the population mean, μ, for given sample standard deviation, s is: ( – (/2; n – 1­) (s/√n), + (/2; n – 1) (s/√n)).

Here, n is the sample size, is the sample mean, and /2; n – 1 is the critical value of the t-distribution with (n – 1) degrees of freedom, above...

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