Garden, Inc., a qualifying § 501(c)(3) organization, incurs lobbying expenditures of $210,000 during the taxable year. Exempt purpose expenditures are $900,000. If Garden makes the election under § 501(h) to make lobbying expenditures on a limited basis, its tax liability resulting from the lobbying expenditures is:

The P-Value ≅0 (zero).

Explanation:

From the given data we have

Regular coffee drinkers sample size = n1 = 50

Decaffeinated-coffee drinkers sample size = n2= 40

Regular coffee drinkers sample mean= x1 = 4.35

Decaffeinated-coffee drinkers sample mean = x2= 5.12

Regular coffee drinkers population standard deviation = σ1 = 1.2

Decaffeinated-coffee drinkers population standard deviation = σ2= 1.36

1) Formulate null and alternate hypothesis

H0: u1≥ u2 Ha: u1 < u2

The null hypothesis is that the mean of the regular coffee drinkers is greater or equal to the mean of decaffeinated-coffee drinkers

against the claim

the mean daily consumption of regular-coffee drinkers is less than that of decaffeinated-coffee drinkers.

2) The test statistic is

z= x1-x2/ sqrt( σ1 ²/n1 + σ2²/n2)

Putting the values

z = 4.35- 5.12/ sqrt( 1.44/50 + 1.8496/40)

z= -5.44

3) The significance level is 0.01

The critical region is Z < -2.33

4) Since the calculated value of z= -5.44 is less than the z ∝= -2.33 we reject H0.

5) the P-value can be calculated using the calculator.

The P-Value is < 0.00001.

P= 0

Which means that the claim is accepted that the mean of the regular coffee drinkers is less than the mean of decaffeinated-coffee drinkers.