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A. <.or=50 B. >220 C.<.50 D.>.50 2.State your decision rule. Fail to Reject H0 i

ID: 3154454 • Letter: A

Question

A. <.or=50

B. >220

C.<.50

D.>.50

2.State your decision rule. Fail to Reject H0 if Z is?

A. >2.33

B. <2.58

C. <-1.65

D. <2.33

3.What is the calculated Z value? Z=?

A.2.0

B.1.20

C.20

D.6.25

4.What is your statistical decision on H0?

A.Fail to reject Ho

B.cannot determine

C.reject Ho

D.need to conduct another sample

5.At the 1 % level of significance does the evidence indicate that more than 50% of the homes in the city are watching American Idol?

A. no

B.cannot determine

C. yes

D. less than 50% are watching

6. What is the p-value?

A. .1151

B.approximately 0

C..1056

D..0228

Explanation / Answer

Here we have to test the hypothesis that,

H0 : p = 50% = 0.5 VS H1 : p > 0.5

where p is population proportion.

q = 1 - p = 1 - 0.5 = 0.5

1. State the alternate hypothesis.

D) H1:>.50

2.State your decision rule. Fail to Reject H0 if Z is?

B. <2.58

Here 2.58 is critical value found by using EXCEL.

syntax is,

=NORMSINV(probability)

where probability = 1 - a/2

a = 0.01

We get critical value is 2.58.

Reject H0 at 1% level of significance if Z > 2.58.

And fail to reject H0 at 1% level of significance if Z < 2.58.

3.What is the calculated Z value?

Test statistic Z = (p^ - p) / sqrt((pq)/n)

p^ is sample proportion = x/n = 220 / 400 = 0.55

Z = (0.55 - 0.5) / sqrt((0.5*0.5) / 220) = 0.05 / 0.025 = 2

Option a) is correct.

4.What is your statistical decision on H0?

Z < 2.58

fail to reject H0 at 1% level of significance.

Option a) is correct.

5.At the 1 % level of significance does the evidence indicate that more than 50% of the homes in the city are watching American Idol?

No

At the 1% level of significance there is not sufficient evidence to say that more than 50% of the homes in the city are watching American Idol.

6. What is the p-value?

P-value we can find by using EXCEL.

syntax is,

=1 - NORMSDIST(z) the test is right sided

where z is test statistic value.

P-value = 0.0228

D..0228

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