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A survey was taken of randomly selected​ Americans, age 65 and​ older, which found that 401 of 1020 men and 536 of 1059 women suffered from some form of arthritis. ​a) Are the assumptions and conditions necessary for inference​ satisfied? Explain. ​b) Create a​ 95% confidence interval for the difference in the proportions of senior men and women who have this disease. ​c) Interpret your interval in this context. ​d) Does this suggest that arthritis is more likely to afflict women than​ men? Explain.

Respuesta :

Answer:

a) yes The assumptions and conditions necessary for inference.

b)The 95% of confidence intervals for p₁-p₂ is determined  by

(-0.15523,-0.07057)

c) The sample proportion of men = 0.3931 = 39%

The sample proportion of women = 0.5061 = 50%

we observe that arthritis is more likely to afflict women than​ men

Step-by-step explanation:

Explanation:-

a)  yes The assumptions and conditions necessary for inference.

b)

Given A survey was taken of randomly selected​ Americans, age 65 and​ older, which found that 401 of 1020 men suffered from some form of arthritis.

Given first sample size 'n₁' = 1020

First sample proportion

                                  'p₁' =  [tex]\frac{x}{n} = \frac{401}{1020} = 0.3931[/tex]

                                    q₁ = 1-p₁ = 1-0.3931 =0.6069

Given A survey was taken of randomly selected​ Americans, age 65 and​ older, which found that 536 of 1059 men suffered from some form of arthritis

Given second sample size 'n₂' = 1059

Second sample proportion

                                    [tex]p_{2} = \frac{x}{n} = \frac{536}{1059} = 0.5061[/tex]

                                    q = 1-p =1-0.5061 =0.4939

Step(ii):-

The 95% of confidence intervals for p₁-p₂ is determined  by

[tex][p_{1} - p_{2} - Z_{\alpha } S.E(p_{1} - p_{2}) ,p_{1} - p_{2} + Z_{\alpha } S.E(p_{1} - p_{2})][/tex]

where

        [tex]S.E (p_{1} -p_{2} ) =\sqrt{\frac{p_{1} q_{1} }{n_{1} } +\frac{p_{2} q_{2} }{n_{2} } }[/tex]

       [tex]S.E (p_{1} -p_{2} ) =\sqrt{\frac{0.3931 X0.6069 }{1020 } +\frac{0.506X0.494 }{1059} }[/tex]

       [tex]S.E (p_{1} -p_{2} ) = \sqrt{0.0004699}[/tex]

       [tex]S.E (p_{1} -p_{2} ) = 0.0216[/tex]

Step(iii):-

The 95% of confidence intervals for p₁-p₂ is determined  by

[tex][p_{1} - p_{2} - Z_{\alpha } S.E(p_{1} - p_{2}) ,p_{1} - p_{2} + Z_{\alpha } S.E(p_{1} - p_{2})][/tex]

(0.3931-0.506 - 1.96 ×0.0216 , 0.3931-0.506 +1.96 ×0.0216)

(-0.1129-0.04233, -0.1129+0.04233)

(-0.15523,-0.07057)

Final answer:-

95% confidence interval for the difference in the proportions of senior men and women who have this disease.

(-0.15523,-0.07057)

c)

The sample proportion of men = 0.3931 = 39%

The sample proportion of women = 0.5061 = 50%

we observe that arthritis is more likely to afflict women than​ men