STATS 10 MCQ

CAN ANY OF YOU DO IT PERFECT , I WANT THAT ANSWERS BE FEEDED IN TOP FREEN COLUMN

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>ONLINE TEST

ST NAME:

2
6
A

o not reject a false alternative.

Type I:

Type II:

Type I:

Type II: Reject a true null hypothesis.

D Type I:

Type II: Reject a true null hypothesis.

. Then you should,

minutes to and from work in a typical day. In a sample of

employed adults in Indianapolis metropolitan area (IMA), the average commuting time to work was

minutes, with

minutes. Is there enough evidence, at α = 0.05, to conclude that the average commuting time in IMA is less than the national average?

B 1.374

Conclude the average commuting time in IMA is less than the national average.

D 2.292 DNR H₀ Conclude the average commuting time in IMA is not less than the national average.
A

for men. In random sample n =

Indiana married men, the average age at marriage was x̅ =

with a standard deviation of s =

years. Does this sample provide a significant evidence that the average age at marriage among Indiana men is below the national average?

|TS| = _______.
A

Reject H₀.

B 1.54

Reject H₀. Conclude that the average age at marriage for Indiana men is below the national average

D 2.35 DNR H₀. Conclude that the average age at marriage for Indiana men is not below the national average

milliliters (ml). A random sample of n = 24 bottles provided a sample mean of x̅ =

with a standard deviation of s =

. Does the sample provide significant evidence that the mean fill exceeds the expected fill? Compute the test statistic (TS) and the critical value at the 5% level of significance.

_______.

Reject H₀.

B 2.107 DNR H₀.

Reject H₀. Conclude the mean fill is greater than the expected fill.

D 1.667 DNR H₀. Conclude the mean fill is not greater than the expected fill.

adults in Indiana, the proportion who were obese was

. Does the sample provide significant evidence that the obesity rate in Indiana is higher than the national rate? State the null and alternative hypotheses and compute the test statistic (TS) and the probability value.

DNR H₀.

B 0.0307 Reject H₀.

DNR H₀. Conclude the Indiana obesity rate is not higher than the national rate.

D 0.0566 Reject H₀. Conclude the Indiana obesity rate is higher than the national rate.

of IUPUI male students the proportion who smoked was p̅ =

. Does the sample provide significant evidence that the proportion of male IUPUI students who smoke is different from the national rate? State the null and alternative hypotheses and compute the test statistic.

DNR H₀.

B 1.84 Reject H₀.

DNR H₀. Conclude that proportion IUPUI male students who smoke is not different from the national rate.

D 2.07 Reject H₀. Conclude that proportion IUPUI male students who smoke is different from the national rate.

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	1)		Which of the following statements about Type I and Type II errors is correct:
							Type I:		Reject a true alternative hypothesis.						Type II:								D
																	B				Reject a true null hypothesis.	Do not reject a false null hypothesis.
																		C		Do not reject a false null hypothesis.
	Reject a false null hypothesis.
	2)		In a test of hypothesis, the symbol α represents:
	The level of significance, or the probability of rejecting a true H₀.
	The level of significance, or the probability of not rejecting a false H₀.
	The power of the test, or the probability of rejecting a false H₀.
	The level of significance, or the probability of committing a Type II error.
	3)	Suppose in a test of hypothesis the probability value (p-value) is		0.0694
	not reject H₀ at								α =	0.10
	reject H₀ at α =							0.05
	reject H₀ at α = 0.10, but not reject H₀ at α = 0.05.
	reject H₀ at α = 0.05, but not reject H₀ at α = 0.10.
	4)		American workers spend an average of	46					n =	81		x̅ =	40.5		s =	21.6
	To perform a hypothesis test, the null and alternative hypotheses for this test are:
	H₀: μ > 46		H₁: μ ≤ 46
	H₀: μ < 46		H₁: μ ≥ 46
		H₀: μ ≤				H₁: μ >
			H₀: μ ≥				H₁: μ <
	5)	For the previous question, the test statistic is	\|TS\| = _______.
			1.374												Reject H₀		Conclude the average commuting time in IMA is less than the national average.
						DNR H₀				Conclude the average commuting time in IMA is not less than the national average.
					2.292									Reject H₀.
	6)		In the previous question the, p-value ≈ ______. (Use z to find the approximate value.)
	0.011
	0.0	24
	0.038
	0.045
	7)		The average age at marriage in the United State is			26.8	110	25.4	6.25
	State the null and alternative hypotheses and compute the test statistic. For the decision rule use α = 0.05. The test statistic is,
			1.54			Conclude that the average age at marriage for Indiana men is below the national average
														DNR H₀.		Conclude that the average age at marriage for Indiana men is not below the national average
					2.35
	8)	The expected fill of bottles of a beverage filled in a bottling plant is			750	750.67	1.971				TS =
			2.107				Conclude the mean fill is greater than the expected fill.
			Conclude the mean fill is not greater than the expected fill.
					1.667
	9)		The national obesity statistics indicates that 25.6% of adults in the United States are obese. In a random sample of n =	985		p̅ =	0.282
	Use a 5% level of significance. P-value = ______.
					0.0307				Conclude the Indiana obesity rate is not higher than the national rate.
			Conclude the Indiana obesity rate is higher than the national rate.
			0.0566
	10)		The national tobacco use statistics indicates that 23.1% of male population smoke cigarettes. In a sample of n =	825	0.204
	Use the 5% level of significance to determine the critical value. TS = ______.
					1.84				Conclude that proportion IUPUI male students who smoke is not different from the national rate.
			Conclude that proportion IUPUI male students who smoke is different from the national rate.
			2.07

K

0 1) Which of the following statements about Type I and Type II errors is correct:
B 0 A Type I: Reject a true alternative hypothesis. Type II:

A 0 B Type I:

Type II: Do not reject a false null hypothesis.

C 0 C Type I: Do not reject a false null hypothesis. Type II: Reject a true null hypothesis.
D 0 D Type I: Reject a false null hypothesis. Type II: Reject a true null hypothesis.
C 0
A 0 2) In a test of hypothesis, the symbol α represents:
C 0 A The level of significance, or the probability of rejecting a true H₀.
D 0 B The level of significance, or the probability of not rejecting a false H₀.
B 0 C The power of the test, or the probability of rejecting a false H₀.
A 0 D The level of significance, or the probability of committing a Type II error.
3)

A not reject H₀ at α = 0.10.
B reject H₀ at α = 0.05.
C reject H₀ at α = 0.10, but not reject H₀ at α = 0.05.
D reject H₀ at α = 0.05, but not reject H₀ at α = 0.10.

0.0694 < α = 0.10 Reject H₀

p-value = 0.0694 > α = 0.05 DNR H₀

4) American workers spend an average of 46 minutes to and from work in a typical day. In a sample of n = 81 employed adults in Indianapolis metropolitan area (IMA), the average commuting time to work was x̅ = 40.5 minutes, with s = 21.6 minutes. Is there enough evidence, at α = 0.05, to conclude that the average commuting time in IMA is less than the national average?

To perform a hypothesis test, the null and alternative hypotheses for this test are:
A H₀: μ > 46 H₁: μ ≤ 46
B H₀: μ < 46 H₁: μ ≥ 46 C H₀: μ ≤ 46 H₁: μ > 46
D H₀: μ ≥ 46 H₁: μ < 46

A 1.374 Reject H₀. Conclude the average commuting time in IMA is less than the national average.
B 1.374 DNR H₀ Conclude the average commuting time in IMA is not less than the national average.
C 2.292 Reject H₀. Conclude the average commuting time in IMA is less than the national average.
D 2.292 DNR H₀ Conclude the average commuting time in IMA is not less than the national average.

H₀: μ ≥ 46 H₁: μ < 46 α = 0.05

x̅ = 40.5
s = 21.6
n = 81

2.292

tα,(n − 1) =

TS = 2.292 > CV = 1.664 Reject H₀

6) In the previous question the, p-value ≈ ______. (Use z to find the approximate value.)
A 0.011
B 0.024
C 0.038
D 0.045

7) The average age at marriage in the United State is 26.8 for men. In random sample n = 110 Indiana married men, the average age at marriage was x̅ = 25.4 with a standard deviation of s = 6.25 years. Does this sample provide a significant evidence that the average age at marriage among Indiana men is below the national average?

State the null and alternative hypotheses and compute the test statistic. For the decision rule use α = 0.05. The test statistic is,

TS = _______.

A 1.54 Reject H₀. Conclude that the average age at marriage for Indiana men is below the national average
B 1.54 DNR H₀. Conclude that the average age at marriage for Indiana men is not below the national average

C 2.35 Reject H₀.

D 2.35 DNR H₀.

H₀: μ ≥ 26.8 H₁: μ < 26.8 α = 0.05 µ₀ = 26.8
x̅ = 25.4
s = 6.25
n = 110
se(x̅) = s/√n =

TS = |x̅ −µ₀|/se(x̅) = 2.35

Decision Rule: Reject H₀ if TS > CV

TS = 2.35 > CV = 1.64 Reject H₀

A 2.107 Reject H₀. Conclude the mean fill is greater than the expected fill.
B 2.107 DNR H₀. Conclude the mean fill is not greater than the expected fill.
C 1.667 Reject H₀. Conclude the mean fill is greater than the expected fill.
D 1.667 DNR H₀. Conclude the mean fill is not greater than the expected fill.

H₀: μ ≤ 750 H₁: μ > 750 α = 0.05
µ₀ = 750
x̅ = 750.67
s = 1.971
n = 24
se(x̅) = s/√n =

TS = |x̅ −µ₀|/se(x̅) = 1.667

Decision Rule: Reject H₀ if TS > CV

TS = 1.667 < CV = 1.714 DNR H₀

9) The national obesity statistics indicates that 25.6% of adults in the United States are obese. In a random sample of n = 985 adults in Indiana, the proportion who were obese was p̅ = 0.282. Does the sample provide significant evidence that the obesity rate in Indiana is higher than the national rate? State the null and alternative hypotheses and compute the test statistic (TS) and the probability value.

Use a 5% level of significance. P-value = ______.
A 0.0307 DNR H₀. Conclude the Indiana obesity rate is not higher than the national rate.
B 0.0307 Reject H₀. Conclude the Indiana obesity rate is higher than the national rate.
C 0.0566 DNR H₀. Conclude the Indiana obesity rate is not higher than the national rate.
D 0.0566 Reject H₀. Conclude the Indiana obesity rate is higher than the national rate.

H₁: μ > 0.256 α = 0.05

0.256

n = 985
p̅ = 0.282

0.0307

Decision Rule: Reject H₀ if p-value < α

p-value = 0.0307 < α = 0.05 Reject H₀

10) The national tobacco use statistics indicates that 23.1% of male population smoke cigarettes. In a sample of n = 825 of IUPUI male students the proportion who smoked was p̅ = 0.204. Does the sample provide significant evidence that the proportion of male IUPUI students who smoke is different from the national rate? State the null and alternative hypotheses and compute the test statistic.

Use the 5% level of significance to determine the critical value. TS = ______.
A 1.84 DNR H₀. Conclude that proportion IUPUI male students who smoke is not different from the national rate.

B 1.84 Reject H₀. Conclude that proportion IUPUI male students who smoke is different from the national rate.
C 2.07 DNR H₀. Conclude that proportion IUPUI male students who smoke is not different from the national rate.

D 2.07 Reject H₀. Conclude that proportion IUPUI male students who smoke is different from the national rate.

0.231 α = 0.05

π₀ = 0.231
n = 825
se(p̅) = √π₀(1 − π₀) =

p̅ = 0.204
TS = |p̅ −π₀|/se(p̅) = 1.84

Decision Rule: Reject H₀ if TS > CV

TS = 1.84 < CV = 1.96 DNR H₀

Do not reject a false alternative.
Reject a true null hypothesis.
Suppose in a test of hypothesis the probability value is 0.0694. Then you should,
	Decision Rule: Reject H₀ if p-value < α
		p-value =
We are testing the hypothesis that the average commuting time in IMA is “less than” (<) the national average. This is a strict inequality and it must be the alternative hypothesis.
For the previous question, the test statistic is	TS = _______.
		µ₀ =
		se(x̅) = s/√n =	2.400
		TS = \|x̅ −µ₀\|/se(x̅) =
				CV =	1.664
			Decision Rule: Reject H₀ if TS > CV
p-value = P(t > TS)
p-value = P(tdf = 80 > 2.292) ≈ P(z > 2.29) =	0.0110
Conclude that the average age at marriage for Indiana me is below the national average
Conclude that the average age at marriage for Indiana me is not below the national average
0.596
CV = zα =		1.64
The expected fill of bottles of a beverage filled in a bottling plant is 750 milliliters (ml). A random sample of n = 24 bottles provided a sample mean of x̅ = 750.67 with a standard deviation of s = 1.971. Does the sample provide significant evidence that the mean fill exceeds the expected fill? Compute the test statistic (TS) and the critical value at the 5% level of significance. TS = _______.
0.402
CV = tα,(n − 1) =		1.714
H₀: π ≤			0.256
	π₀ =
	se(p̅) = √π₀(1 − π₀) =	0.0139
	TS = \|p̅ −π₀\|/se(p̅) =	1.87
p-value = P(z > TS) =
H₀: π =			0.231	H₁: μ ≠
0.0147
CV = zα/2 =		1.96

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