QUESTION 4 A 2003 New York Times CBS news poll sampled 523 adults who were plann

Question

QUESTION 4 A 2003 New York Times CBS news poll sampled 523 adults who were planning a vacation during the next 6 months and found that 141 were expecting to travel by airplane. A similar survey question in May 1993 New York Times CBS news poll found that of 477 adults who were planning to take a vacation in the next 6 months, 81 were planning to travel by airplane. State the hypotheses that can be used to determine whether a significant change occurred in the population proportion planning to travel by airplane over 10 years.

QUESTION 5

If ?=.03 do you accept or reject the null hypothesis?

Accept

Reject

QUESTION 6

Develop a 97% confidence interval for the data in problem 4.

[.02, .14]

[.06, .18]

[.04, .16]

Cash

the data needed to solve problem 6 is above , pelease solve all 3 questions using excel and circle final answers

74

Explanation / Answer

H0 : p1 = p2
Ha : p1 not equals to p2

Test statistics;

p1 = 141/523 = 0.2696 , n1 = 523
p2 = 81/477 = 0.1698, n2 = 477

p = (p1 * n1 + p2 * n2) / (n1 + n2)
= (0.2696 * 523 + 0.1698 * 477) / ( 523 + 477)
= 0.2220

SE = sqrt{ p * ( 1 - p ) * [ (1/n1) + (1/n2) ] }
= sqrt( 0.222 * (1-0.222) *((1/523) + ( 1/477)))
= 0.0263

z = (p1 - p2) / SE
= ( 0.2696 - 0.1698)/0.0263
= 3.7947

p value = 0.0001

Reject the null hypothesis

6)

z value at 97% = 2.1701

CI = ( p1 -p2) +/- z *sqrt(p1 *(1-p1)/n1 + p2 *(1-p2)/n2)
  
= ( 0.2696 - 0.1698) +/- 2.1701 * sqrt(0.2696 *(1-0.2696)/523 + 0.1698 *(1-0.1698)/477)
= ( 0.0435,0.15610

The 97% CI is (0.04 , 0.16)

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