A recent tax hike has been proposed to pay for new school facilities. Members of the school board want to know if the tax will pass a vote by the people. They conduct interviews with 799 randomly selected likely voters and found 431 of the people said they would vote in favor of the tax. They conduct a hypothesis test of H:p=0.5 against H:p > 0.5 to determine if the tax is expected to pass.

Part 1: Input the test statistic.

Part 2: Input the p-value.

Solution :

Given that,

 = 0.5

1 -  = 0.5

n =799

x =431

Level of significance =  = 0.05

Point estimate = sample proportion =  = x / n = 431 /799 = 0.5394

This a right- tailed test.

The null and alternative hypothesis is,

Ho: p = 0.5

Ha: p > 0.5

part1:

Test statistics

z = ( -  ) /  *(1-) / n

= ( 0.5394 - 0.5) /  ((0.5*0.5) / 799)

= 0.0394 / 0.0177

= 2.23

Test statistic = 2.23

part2:

P-value = P(Z>z)

= 1 - P(Z <z )

= 1- P(Z < 2.23)

= 1 - 0.9871

=0.0129

The p-value is p =0.0129, and since p = 0.0129 < 0.05, it is concluded that the null hypothesis is rejected.

Conclusion:

There is sufficient evidence to conclude that the tax is expected to pass.