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Orlov [11]
3 years ago
9

Individuals filing federal income tax returns prior to March 31 received an average refund of $1056. 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 researcher’s 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 σσ = $1600 may be assumed. What is the p-value?
c. At αα = .05, what is your conclusion?
d. Repeat the preceding hypothesis test using the critical value approach.
Mathematics
1 answer:
Sholpan [36]3 years ago
6 0

Answer:

Step-by-step explanation:

(A) Develop appropriate hypotheses such that the rejection of H0 will support the researcher's contention.

What is the researcher's contention?

The researcher's contention is that late filers receive lower refunds than early filers. Hence, the researcher's contention will be the alternative hypothesis; since the rejection of H0 will imply the acceptance of Ha (the alternative hypothesis).

To this regard, the appropriate hypotheses developed are:

H0: Late filers receive HIGHER refunds than early filers

H1: Late filers receive lower refunds than early filers

The rejection of H0 will lead to the acceptance of H1; which is in line with the researcher's contention/claim.

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I toss an unfair coin 12 times. This coin is 65% likely to show up heads. Calculate the probability of the following.
mars1129 [50]

Answer:

a. 0.0368

b. 0.99992131

c. 0.2039

d. 0.0048

e. 0.6533

Step-by-step explanation:

Let the probability of obtaining a head be p = 65% = 13/20 = 0.65. The probability of not obtaining a head is q = 1 - p = 1 -13/20 = 7/20 = 0.35

Since this is a binomial probability, we use a binomial probability.

a. The probability of obtaining 11 heads is ¹²C₁₁p¹¹q¹ = 12 × (0.65)¹¹(0.35) = 0.0368

b. Probability of 2 or more heads P(x ≥ 2) is

P(x ≥ 2) = 1 - P(x ≤ 1)

Now P(x ≤ 1) = P(0) + P(1)

= ¹²C₀p⁰q¹² + ¹²C₁p¹q¹¹

= (0.65)⁰(0.35)¹² + 12(0.65)¹(0.35)¹¹

= 0.000003379 + 0.00007531

= 0.0007869

P(x ≥ 2) = 1 - P(x ≤ 1)

= 1 - 0.00007869

= 0.99992131

c. The probability of obtaining 7 heads is ¹²C₇p⁷q⁵ = 792(0.65)⁷(0.35)⁵ = 0.2039

d. The probability of obtaining 7 heads is ¹²C₉q⁹p³ = 220(0.65)³(0.35)⁹ = 0.0048

e. Probability of 8 heads or less P(x ≤ 8) = ¹²C₀p⁰q¹² + ¹²C₁p¹q¹¹ + ¹²C₂p²q¹⁰ + ¹²C₃p³q⁹ + ¹²C₄p⁴q⁸ + ¹²C₅p⁵q⁷ + ¹²C₆p⁶q⁶ + ¹²C₇p⁷q⁵ + ¹²C₈p⁸q⁴

= = ¹²C₀(0.65)⁰(0.35)¹² + ¹²C₁(0.65)¹(0.35)¹¹ + ¹²C₂(0.65)²(0.35)¹⁰ + ¹²C₃(0.65)³(0.35)⁹ + ¹²C₄(0.65)⁴(0.35)⁸ + ¹²C₅(0.65)⁵(0.35)⁷ + ¹²C₆(0.65)⁶(0.35)⁶ + ¹²C₇(0.65)⁷(0.35)⁵ + ¹²C₈(0.65)⁸(0.35)⁴

= 0.000003379 + 0.00007531 + 0.0007692 + 0.004762 + 0.01990 + 0.05912 + 0.1281 + 0.2039 + 0.2367

= 0.6533

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3 years ago
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-14/15y+3/21=-14/21

Move 3/21 to the other side. Sign changes from +3/21 to -3/21.

-14/15y+3/21-3/21=-14/21-3/21

-14/15y=-14/21-3/21

-14/21-3/21=-17/21

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Multiply both sides by -15/14

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Cross out 15 and 15, divide by 15 then becomes 1

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1*1*y=y

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Answer: c. y=85/98

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What is 2 1/8 × 2 3/5 in the simplest form
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Select the following series as finite or infinite.
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Infinite because it keeps going

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