document.write( "Question 1182685: Assume that we want to construct a confidence interval. Do one of the following, as appropriate: (a) find the critical value tα/2, (b) find the critical value zα/2, or (c) state that neither the normal distribution nor the t distribution applies.\r
\n" );
document.write( "\n" );
document.write( "Here are summary statistics for randomly selected weights of newborn girls: n = 187, x̄ = 28.1 hg, s = 6.1 hg. The confidence level is 95%.\r
\n" );
document.write( "\n" );
document.write( "Select the correct choice below and, if necessary, fill in the blank to complete your choice.\r
\n" );
document.write( "\n" );
document.write( "A. tα/2 = __
\n" );
document.write( "(Round to two decimal places as needed.)\r
\n" );
document.write( "\n" );
document.write( "B. zα/2 = __
\n" );
document.write( "(Round to two decimal places as needed.)\r
\n" );
document.write( "\n" );
document.write( "C. Neither the normal distribution nor the t distribution applies. \n" );
document.write( "
Algebra.Com's Answer #812822 by Boreal(15235)![]() ![]() You can put this solution on YOUR website! It is a relatively large sample size, and if one believes in the rule of 30 regardless, a z-test would be used. However, I prefer using a t-test since the sd used is from the sample. The difference is <1% with a sample size this large. While the population weight is skewed somewhat, it is not likely to such a degree that a random sample of 187 could not be construed to be normally distributed. \n" ); document.write( "t(alpha/2)=1.97 \n" ); document.write( "z(alpha/2)=1.96 \n" ); document.write( "If you've not studied the rule of 30 for sample size, use t. \n" ); document.write( "If you were taught the rule of 30 means you can use z, use it. \n" ); document.write( "But notice the small difference between the two with this sample size. \n" ); document.write( " |