SOLUTION: 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 Coa

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.
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
east coast population mean is 7.58 with standard deviation of .42
west coast sample mean is 7.65 with sample size = 81.

the standard error for the test is population standard deviation divided by sample size = .42 / sqrt(81) = .046666667 rounded to 9 decimal places.

the z-score is (sample mean minus population mean) / standard error.
this is equal to (7.65 - 7.58) / .046666667 = 1.5

area to the left of this z-score is .9331927713.
area to the right of this z-score is .0668072287.

the p-value for the test is the area to the right of this z-score.

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.

at 95% confidence interval, the critical p-value would be .05.
at 90% confidence interval, the critical p-value would be .10.

at .0668072287 p-value for the test, the results would be significant at 90 confidence interval but not significant at 95% confidence interval.

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.

since you have the population standard deviation, and since the sample size is greater than 30, a z-test would be indicated.

the standard error is equal to the population standard deviation divided by the square root of the sample size.
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.
rounding to 4 decimal places would have probably been good enough, but i just used what the calculator displayed.
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.
the p-value, in that case, would have been .0669 rounded to 4 decimal places.

either way, the result would have been the same.
the results were significant at 10% confidence level, but not significant at 5% significance level.

since the test was whether the sample mean was significantly greater than the population mean, the one tailed confidence interval was indicated.

here's a reference.

https://www.statisticshowto.com/one-sample-z-test/

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