document.write( "Question 1182978: Assume that we want to construct a confidence interval. (a) Find the critical value t α/2, (b) find the critical value z α/2 or (c) state that neither the normal distribution nor the t distribution applies.\r
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document.write( "Here are summary statistics for randomly selected weights of newborn girls: n = 169, x = 29.2 hg, s = 7.3 hg. The confidence level is 95% \n" );
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Algebra.Com's Answer #813202 by Boreal(15235)![]() ![]() You can put this solution on YOUR website! I would use a t-test since the sd of the sample is used as a variability estimator. It is true that n is > 30, so the difference between z and t is < 1%. Furthermore, given the ease of calculators today, dealing with a t-table is essentially the same as dealing with a z-table. \r \n" ); document.write( "\n" ); document.write( "The critical z-value (0.975) is 1.96; the t-value (0.975, df=168) is 1.97. My answer would be a, but some would say use z. The difference is small.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "--- \n" ); document.write( "the half-interval for the interval is t*s/sqrt(n) \n" ); document.write( "=1.97*7.3/13 \n" ); document.write( " \n" ); document.write( " |