Question

A tax preparer claims that the mean income tax in Texas for the year of 2018 is at least $25000 per household. A random sample of 35 households’ income taxes of Year 2018 in Texas shows the sample’s mean x = $24800. Assume the population standard deviation  = $1800. Test the claim of the tax preparer at  = 0.01.

Answer 1

Let X denote the income tax in Texas for the year of 2018

Claim is that X is atleast $25000

Null hypothesis, H​​​​​​0 : ≥ $25000

Alternative hypothesis, H​​​​​​a : < $25000

Sample mean = $24800

Standard deviation = $1800

n = 35

Thus, the test statistic = (24800 - 25000)/(1800/√35)

= -0.657

The P-value = P(Z < -0.657)

= 0.2535

Since the P-value is greater than the significance level, we fail to reject the Null hypothesis.

Thus, the claim that the income tax is atleast $25000 cannot be denied