document.write( "Question 1179918: A recent survey indicated that the average amount spent for breakfast by business managers was $7.58 with a standard deviation of $0.42. It was felt that breakfasts on the West Coast were higher than $7.58. A sample of 81 business managers on the West Coast had an average breakfast cost of $7.65. Find the P-value for the test. \n" ); document.write( "
Algebra.Com's Answer #809559 by Theo(13342)\"\" \"About 
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east coast population mean is 7.58 with standard deviation of .42
\n" ); document.write( "west coast sample mean is 7.65 with sample size = 81.\r
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\n" ); document.write( "\n" ); document.write( "the standard error for the test is population standard deviation divided by sample size = .42 / sqrt(81) = .046666667 rounded to 9 decimal places.\r
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\n" ); document.write( "\n" ); document.write( "the z-score is (sample mean minus population mean) / standard error.
\n" ); document.write( "this is equal to (7.65 - 7.58) / .046666667 = 1.5\r
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\n" ); document.write( "\n" ); document.write( "area to the left of this z-score is .9331927713.
\n" ); document.write( "area to the right of this z-score is .0668072287.\r
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\n" ); document.write( "\n" ); document.write( "the p-value for the test is the area to the right of this z-score.\r
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\n" ); document.write( "\n" ); document.write( "since the test was whether the west coast average breakfast cost was higher than the east coast breakfast cost, then a one tailed confidence interval would be used to determine the significance of the test.\r
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\n" ); document.write( "\n" ); document.write( "at 95% confidence interval, the critical p-value would be .05.
\n" ); document.write( "at 90% confidence interval, the critical p-value would be .10.\r
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\n" ); document.write( "\n" ); document.write( "at .0668072287 p-value for the test, the results would be significant at 90 confidence interval but not significant at 95% confidence interval.\r
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\n" ); document.write( "\n" ); document.write( "the most important calculation you would make would be to determine what the standard error for the test would be and whether or not a z-test or a t-test is indicated.\r
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\n" ); document.write( "\n" ); document.write( "since you have the population standard deviation, and since the sample size is greater than 30, a z-test would be indicated.\r
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\n" ); document.write( "\n" ); document.write( "the standard error is equal to the population standard deviation divided by the square root of the sample size.
\n" ); document.write( "that's why the standard error of the test was calculated to be .42/sqrt(81) = .046666667 rounded to the number of decimal places that the calculator was able to display.
\n" ); document.write( "rounding to 4 decimal places would have probably been good enough, but i just used what the calculator displayed.
\n" ); document.write( "if i had rounded to 4 decimal places, then the z-score would have been (7.65 - 7.58) / .0467 = 1.498929336 which i would have probably rounded to 1.499.
\n" ); document.write( "the p-value, in that case, would have been .0669 rounded to 4 decimal places.\r
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\n" ); document.write( "\n" ); document.write( "either way, the result would have been the same.
\n" ); document.write( "the results were significant at 10% confidence level, but not significant at 5% significance level.\r
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\n" ); document.write( "\n" ); document.write( "since the test was whether the sample mean was significantly greater than the population mean, the one tailed confidence interval was indicated.\r
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\n" ); document.write( "\n" ); document.write( "here's a reference.\r
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\n" ); document.write( "\n" ); document.write( "https://www.statisticshowto.com/one-sample-z-test/\r
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