Question 1184879
population mean assumed to be 23% = .23
alaska had 321 complaints out of 1432.
the proportion for alaska is 321/1432 = .22416.
this is the sample mean.
the sample size is 1432.


the standard error is equal to sqrt(population mean * (1 - population mean) / sample size) = sqrt(.23 * (1-.23) / 1432) = .0111208.


z-score is (x - m) / s
x is the sample mean.
m is the population mean
s is the standard error.


z-score = (.22416 - .23) / .0111208 = -.52514.


the area to the left of that z-score is .2997.


the critical z-score at .05 two tailed significance level is equal to -1.96.


the absolute value of the test test z-score is not greater than the absolute value of the critical z-score.


this means the results are not significant and there is not enough evidence to show that the alaska proportion is different from the general population proportion.


also, the critical p-value is .025 and the test p-value is .2997.
this is another indication that the test results are not significant.


the critical z-score and the critical p-value will always agree.
either one is sufficient to determine significance of the test.