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2. A simple random sample of size n is drawn. The sample mean I is found to be 53.1, and the sample standard deviation s is f

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are -> Giren summary statistics x 53.1 and ase M at L20.05 Given that 281 The degrees of freedom difla 81-80 95% confidence i=> le (53.1 I to.0572,28 -> (he) (534132 (830122.04523 X 10 42407865) 53.11 2.04523 x 7.8 5.47722558 al a e) (53.112.91256838@ from as the width of the confidence interval is width a upper limita lower limit width 54.8247 - 9.3753 width = 3.4494 from

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