(-0.0748, 0.15668) (-8.265, 9.2652) (-0.2668, 1.2668) 0.2335 0.0878) (-7.034, 8.

Question

(-0.0748, 0.15668) (-8.265, 9.2652) (-0.2668, 1.2668) 0.2335 0.0878) (-7.034, 8.0342) (-11.9, 12.896) (1.109, -0.491) (0.2161 0.1062) (-0.2072, 0.1142) in 2016, 535 adults aged 18 years old or older were asked i they think that here is some form of lite on other pianets in J (1,.11, the universe or not (Group 1). Of the 535 surveyed, 326 said yes. When the same question was asked in 2006, 385 of the 0.4889) 500 individuals surveyed said yes (Group 2). Construct a 90% condence interval for te dfference between hose in 2006 and 2016 who believe there is life on other planets K (0.83037, 1.5696) How often dok ou go out dancing? This question was asked by a professional survey group on behalt of the National Ats Survey A random sample of 76 single men showed that 26% went ot dreng ocasionally Another fanden sample of 72 single women showed that 22% went out daning ocas rally 11 the propor on of single men who go din ng

Explanation / Answer

Question 1

Solution:

Here, we have to construct the confidence interval for the difference between two population proportions for those in 2006 and 2016 who believe there is life on other planets.

Confidence interval for difference between two population proportions is given as below:

Confidence interval = (P1 – P2) -/+ Z*sqrt[(P1*(1 – P1)/N1) + (P2*(1 – P2)/N2)]

Where, P1 and P2 are sample proportions for first and second groups respectively.

We are given

Confidence level = C = 90% = 0.90

Z = 1.6449 (by using z-table)

X1 = 326, N1 = 535, P1 = X1/N2 = 326/535 = 0.609345794

X2 = 385, N2 = 500, P2 = X2/N2 = 385/500 = 0.77

(P1 – P2) = 0.609345794 - 0.77 = -0.16065421

Confidence interval = (P1 – P2) -/+ Z*sqrt[(P1*(1 – P1)/N1) + (P2*(1 – P2)/N2)]

Confidence interval = (0.609345794 - 0.77) -/+ 1.6449*sqrt[(0.609345794*(1 – 0.609345794)/535) + (0.77*(1 – 0.77)/500)]

Confidence interval = (0.609345794 - 0.77) -/+ 1.6449* 0.0283

Confidence interval = (0.609345794 - 0.77) -/+ 0.0465

Confidence interval =-0.16065421 -/+ 0.0465

Lower limit = -0.16065421 - 0.0465 = -0.2072

Upper limit = -0.16065421 + 0.0465 = -0.1142

Confidence interval = (-0.2072, -0.1142)

Question 2

Here, we have to construct the confidence interval for the difference between two population proportions.

Confidence interval for difference between two population proportions is given as below:

Confidence interval = (P1 – P2) -/+ Z*sqrt[(P1*(1 – P1)/N1) + (P2*(1 – P2)/N2)]

Where, P1 and P2 are sample proportions for first and second groups respectively.

We are given

Confidence level = C = 90% = 0.90

Z = 1.6449 (by using z-table)

N1 = 76, P1 = 0.26, N2 = 72, P = 0.22

Confidence interval = (0.26 – 0.22) -/+ 1.6449*sqrt[(0.26*(1 – 0.26)/76)+(0.22*(1 – 0.22)/72)]

Confidence interval = 0.04 -/+ 1.6449*0.070106435

Confidence interval = 0.04 -/+ 0.115318075

Lower limit = 0.04 – 0.1153 = -0.0753

Upper limit = 0.04 + 0.1153 = 0.1553

Confidence interval = (-0.0753, 0.1553)

Above confidence interval includes the value of zero, so we conclude that there is no any significant difference between the proportion of men and women who went out dancing occasionally.

There is sufficient evidence to conclude that the proportion of men and women who went out dancing occasionally is same.

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