Question

A.

x 9.2.7-T Question Help 12 A study was done on proctored and nonproctored tests. The results are shown in the table. le Proctored Nonproctored Assume that the two samples are independent simple random samples selected from HH1 normally distributed populations, and do not assume that the population standard In 33 31 deviations are equal. Complete parts (a) and (b) below. Use a 0.05 significance level for X 75.47 87.09 both parts. 8 11.57 19.98 a. Test the claim that students taking nonproctored tests get a higher mean score than those taking proctored tests. What are the null and alternative hypotheses? A Ho H1 H2 Hy: Hy <H2 C. Ho H1 H2 Hy: Hy* H2 OB. Ho! H1 H2 H4: Hi > H2 OD. HOE HH2 Hy: HH2 The test statistic, t. is . (Round to two decimal places as needed.) Enter your answer in the answer box and then click Check Answer. A parts 4 remaining Clear All Check Answer

B.

Please follow the steps of hypothesis testing, including identifying the alternative and null hypothesis, calculating the test statistic, finding the p-value, and making a conclusions about the null hypothesis and a final conclusion that addresses the original claim. Use a significance level of 0.10. Is the conclusion affected by whether the significance level is 0.10 or 0.01?

Test Statistic=______ (Round to two decimal places)

P-Value=______ (Round to three decimal places)

Answer choices below:
a) ​Yes, the conclusion is affected by the significance level because H0 is rejected when the significance level is 0.01 but is not rejected when the significance level is 0.10.
b) No, the conclusion is not affected by the significance level because H0 is not rejected regardless of whether a significance level of 0.10 or 0.01 is used.
c) Yes, the conclusion is affected by the significance level because H0 is rejected when the significance level is 0.10 but is not rejected when the significance level is 0.01.
d) ​No, the conclusion is not affected by the significance level because H0 is rejected regardless of whether a significance level of 0.10 or 0.01 is used.

C.

D.

P.S Can you please answer part A-D. I am very sick and I can't think right now. Please help me im begging you. My assignment is due tonight and Ive been stuck in bed all week. Please help. May god bless your soul!

Answer 1

A.

For proctored :

x̅1 = 75.47, s1 = 11.57, n1 = 33

For Nonproctored :

x̅2 = 87.09, s2 = 19.98, n2 = 31

a) Null and Alternative hypothesis:

Ho : µ1 = µ2

H1 : µ1 < µ2

df = ((s1²/n1 + s2²/n2)²)/[(s1²/n1)²/(n1-1) + (s2²/n2)²/(n2-1) ] = 47.4621 = 47

Test statistic:

t = (x̅1 - x̅2)/√(s1²/n1 + s2²/n2) = (75.47 - 87.09)/√(11.57²/33 + 19.98²/31) = -2.82

p-value :

Left tailed p-value = T.DIST(-2.82, 47, 1) = 0.003

Conclusion:

p-value < α, Reject the null hypothesis

Answer C. Reject Ho. There is sufficient evidence to support the claim that students taking Nonproctored test get a higher test score than those taking proctored tests.

b) 95% Confidence interval for the difference :

At α = 0.05 and df = 47, two tailed critical value, t_c = T.INV.2T(0.05, 47) = 2.012

Lower Bound = (x̅1 - x̅2) - t_c*√(s1²/n1 +s2²/n2) = (75.47 - 87.09) - 2.012*√(11.57²/33 + 19.98²/31) = -19.898

Upper Bound = (x̅1 - x̅2) + t_c*√(s1²/n1 +s2²/n2) = (75.47 - 87.09) + 2.012*√(11.57²/33 + 19.98²/31) = -3.342

-19.898 < µ1 - µ2 < -3.342

Yes, because the confidence interval contain only negative values.

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B.

For Recent :

Sample mean using excel function AVERAGE(), x̅1 = 78.8824

Sample standard deviation using excel function STDEV.S, s1 = 13.7153

Sample size, n1 = 17

For Past :

Sample mean using excel function AVERAGE(), x̅2 = 87.5000

Sample standard deviation using excel function STDEV.S, s2 = 7.4039

Sample size, n2 = 12

Null and Alternative hypothesis:

Ho : µ1 = µ2

H1 : µ1 ≠ µ2

df = ((s1²/n1 + s2²/n2)²)/[(s1²/n1)²/(n1-1) + (s2²/n2)²/(n2-1) ] = 25.5931 = 25

Test statistic:

t = (x̅1 - x̅2)/√(s1²/n1 + s2²/n2) = (78.8824 - 87.5)/√(13.7153²/17 + 7.4039²/12) = -2.18

p-value :

Two tailed p-value =T.DIST.2T(-2.18, 25) = 0.039

Answer choices below:

Answer c) Yes, the conclusion is affected by the significance level because H0 is rejected when the significance level is 0.10 but is not rejected when the significance level is 0.01.

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C.

For Archbishops :

Sample mean using excel function AVERAGE(), x̅1 = 13.75

Sample standard deviation using excel function STDEV.S, s1 = 4.9541

Sample size, n1 = 24

For Monarchs :

Sample mean using excel function AVERAGE(), x̅2 = 15.9167

Sample standard deviation using excel function STDEV.S, s2 = 1.4434

Sample size, n2 = 12

Null and Alternative hypothesis:

Ho : µ1 = µ2

H1 : µ1 < µ2

df = ((s1²/n1 + s2²/n2)²)/[(s1²/n1)²/(n1-1) + (s2²/n2)²/(n2-1) ] = 29.6834 = 29

Test statistic:

t = (x̅1 - x̅2)/√(s1²/n1 + s2²/n2) = (13.75 - 15.9167)/√(4.9541²/24 + 1.4434²/12) = -1.98

p-value :

Left tailed p-value = T.DIST(-1.98, 29, 1) = 0.029

Conclusion:

Answer D. Reject the null hypothesis. There is sufficient evidence to support the claim that the archbishops have lower mean longevity than monarchs.

b) 90% Confidence interval for the difference :

At α = 0.1 and df = 29, two tailed critical value, t_c = T.INV.2T(0.1, 29) = 1.699

Lower Bound = (x̅1 - x̅2) - t_c*√(s1²/n1 +s2²/n2) = (13.75 - 15.9167) - 1.699*√(4.9541²/24 + 1.4434²/12) = -4.025

Upper Bound = (x̅1 - x̅2) + t_c*√(s1²/n1 +s2²/n2) = (13.75 - 15.9167) + 1.699*√(4.9541²/24 + 1.4434²/12) = -0.308

-4.025 < µ1 - µ2 < -0.308

Yes, because the confidence interval contain only negative values.

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D.

Actress	Actor	Difference
28	64	-36
27	33	-6
29	32	-3
31	39	-8
39	32	7
27	34	-7
29	47	-18
37	36	1
28	37	-9
32	44	-12

Sample mean of the difference using excel function AVERAGE(), x̅d = -9.1

Sample standard deviation of the difference using excel function STDEV.S(), sd = 11.6662

Sample size, n = 10

Null and Alternative hypothesis:

Ho : µd = 0

H1 : µd < 0

Test statistic:

t = (x̅d)/(sd/√n) = (-9.1)/(11.6662/√10) = -2.47

df = n-1 = 9

Left tailed p-value = T.DIST(-2.4667, 9, 1) = 0.018

Decision:

p-value > α, Do not reject the null hypothesis

Conclusion:

There is not enough evidence to conclude that

Since the p-value is more than the significance level, fail to reject the null hypothesis. There is not sufficient evidence to support the claim that actress are generally younger when they won the award than actors.

b) 99% Confidence interval :

At α = 0.01 and df = n-1 = 9, two tailed critical value, t-crit = T.INV.2T(0.01, 9) = 3.250

Lower Bound = x̅d - t-crit*sd/√n = -9.1 - 3.25 * 11.6662/√10 = -21.1

Upper Bound = x̅d + t-crit*sd/√n = -9.1 + 3.25 * 11.6662/√10 = 2.9

-21.1 < µd < 2.9

Since confidence interval contain zero, fail to reject the null hypothesis.