Consider the population of "last-minute" filers who mail their tax return during the last five days of the income tax period (typically April 10 to April 15). a. A researcher suggests that a reason individuals wait until the last five days is that on average these individuals receive lower refunds than do early filers. Develop appropriate hypotheses such that rejection of H0 will support the researcherâs contention.
b. For a sample of 400 individuals who filed a tax return between April 10 and 15, the sample mean refund was $910. Based on prior experience a population standard deviation of Ï = $1600 may be assumed. What is the p-value?
c. At α = .05, what is your conclusion?
d. Repeat the preceding hypothesis test using the critical value approach.

Answer and Step-by-step explanation:

Solution:

Develop hypothesis:

H₀ : µ ≥ 1056

H₁ < 1056

Given:

n = 400

ẋ = 910

∂ = 1600

a = 0.05

The sampling distribution of sample means µ and standard deviation ∂ / √n.

Z = ẋ-µ /∂ / √n.

  = 910 – 1056 / 1600 /√400.

 = - 1.83

Probability: p = p(z< -1.83) = 0.033

At a = 0.05

If p value is smaller than significance level a, then hypothesis is rejected.

P < 0.05 its means H0 rejected.

Hypothesis test using critical value:

The critical value corresponding to probability of 0.05:

Z = -1.645

The rejection region, then contains all values smaller than -1.645.

If value of test statistics is within the rejection region, then null hypothesis rejected.

-1.83 < -1.645

Rejected H

this doesn't make sense lol

the correct answer is

c. 4

step-by-step explanation:

the given polynomial is

the degree of the polynomial is the highest exponent which is 4.

the -4 is called the leading coefficient.

this polynomial has only one term so it is a monomial.

the correct answer is option c.