document.write( "Question 463673: A ski shop manager claims that the average of the sales for her/his shop is $1800 a day during the winter months. Ten winter days are selected at random, and the mean of the sales is $1830. The standard deviation of the population is $200. Can one reject the claim at α = 0.05? Find the 95% confidence interval of the mean. Does the confidence interval interpretation agree with the hypothesis test results? Explain. Assume that the variable is normally distributed \n" ); document.write( "
Algebra.Com's Answer #317665 by edjones(8007)\"\" \"About 
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H[o]=1800
\n" ); document.write( "\"mu=1800\"
\n" ); document.write( "n=10, m=1830, s=200
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\n" ); document.write( "\"t=%28m-mu%29%2F%28s%2Fsqrt%28n%29%29=%281830-1800%29%2F%28200%2Fsqrt%2810%29%29=.474\"
\n" ); document.write( "t[.025][9]=2.262 (two tailed)
\n" ); document.write( "Since .474<2.262 then we accept the null hypothesis. The ski shop manager is not refuted.
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