Forbes reports that women trust recommendations from Pinterest more than recommendations from any other social network platform (Forbes website, April 10, 2012). But, does trust in Pinterest differ by gender? The following sample data show the number of women and men who stated in a recent sample that they trust recommendations made on Pinterest. a. What is the point estimate of the proportion of women who trust recommendations made on Pinterest (to 2 decimals)? b. What is the point estimate of the proportion of men who trust recommendations made on Pinterest (to 2 decimals)? c. Provide a 95% confidence interval estimate of the difference between the proportion of women and men who trust recommendations made on Pinterest (to 4 decimals). (to) d. Is there enough statistical evidence to show that trust in Pintrest differs by gender (women as population 1)?

Respuesta :

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

[tex]\r p_w = 0.78[/tex]

b

[tex]\r p_m = 0.60[/tex]

c

[tex]0.08 \ to \ 0.28[/tex]

d

Yes, This is because on the interval the two are positive

Step-by-step explanation:

From the table  the point estimate of the proportion of women who trust recommendations made on Pinterest is mathematically evaluated as

          [tex]\r p_w = \frac{117}{150}[/tex]

          [tex]\r p_w = 0.78[/tex]

From the table  the point estimate of the proportion of men who trust recommendations made on Pinterest  is mathematically evaluated as

          [tex]\r p_m = \frac{102}{170}[/tex]

          [tex]\r p_m = 0.60[/tex]

Generally  95% confidence interval estimate of the difference between the proportion of women and men who trust recommendations made on Pinterest is mathematically represented as

         [tex]C= \r p_w - \r p_m \pm Z_{\frac{1-0.95}{2} } \sqrt{ \frac{\r p (1- \r p)}{n_w} + \frac{\r p_m (1- \r p_m)}{n_m} }[/tex]

Where

          [tex]Z_{\frac{1-0.95}{2} }[/tex]  is the is the z-statistic value of level of significance of a two-tail test  

   =>   [tex]Z_{\frac{0.05}{2} } = 1.96[/tex] this is obtained from the the z-table

and  [tex]n_m , n_w[/tex] are the sample size of the men and the women

 So  substituting values from table in the question

        [tex]C= 0.78 -0.60 \pm 1.96 \sqrt{\frac{ 0.78(1-0.78)}{150} + \frac{0.60 (1-0.60)}{170} }[/tex]

       [tex]C= 0.18 \pm 0.10[/tex]

So  the 95% confidence interval is  

    [tex]0.08 \ to \ 0.28[/tex]

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