A partioular poll tracks daly the peicentage of Americans who approve or disappr

Question

A partioular poll tracks daly the peicentage of Americans who approve or disapprove of the perlormance by President 1. Daly resuits are based on random telephone interviews wilth approximately 1000 national adults. The poll reports hat 49% of adults approve of President Te sane po repo ted an approval ating of 50% t President 2 Arews anchor reman sat "President 1 does teenget as much approval as President 2 did. Is there evidence that this dfferenoe is rea? State and test the appropriate hypotheses. Check conditions What are the nul and atemative hypotheses? OD. Hoip 0.50 H P 0.50 Are the assumptions and condtions for the test met? A. No. whie the successFaue Condtion is met, the independence Assumption is not met . No whie the independence Assunpson is me. the SuccessFaure Cordion is not met O C. Yes, the assumptions and conditions are met O D. No neither the Independence Assumption nor the SuccessFalure Condition are met What is the vaue of the test staistic? A The test satis s(Round to two decimal pfaces as needed Click to select your answer

Explanation / Answer

Solution:

Here, we have to use the z test for a population proportion.

The null and alternative hypotheses for this test are given as below:

Null hypothesis: H0: The percentage of adults who approve the President 1 is 50%.

Alternative hypothesis: Ha: The percentage of adults who approve the President 1 is less than 50%.

H0: p = 0.50 vs. Ha: p < 0.50

[Correct answer is C.]

Are the assumption and conditions for the test met?

Yes, the assumption and conditions are met. (np & n(1 – p) 10)

The test statistic formula is given as below:

Z = (P – p) / sqrt(p*q/n)

Where,

P = sample proportion and p = population proportion

n = sample size

q = 1 – p

Here, p = 0.50, so q = 1 – 0.50 = 0.50, n = 1500, P = 0.49

Z = (0.49 - 0.50)/sqrt(0.50*0.50/1500)

Z = -0.77459667

The test statistic is -0.77. (Correct Answer: A)

Critical value = -1.6449 (by using z-table)

P-value = 0.2193 (by using z-table)

P-value = 0.229

P-value > = 0.05

So, we do not reject the null hypothesis

This is not strong evidence that these reports are significantly different.

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