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A study1 conducted in July 2015 examines smartphone ownership by US adults. A random sample of 2001 people were surveyed, and the study shows that 688 of the 989 men own a smartphone and 671 of the 1012 women own a smartphone. We want to test whether the survey results provide evidence of a difference in the proportion owning a smartphone between men and women. Let group 1 be US men and let group 2 be US women.

(a) State the null and alternative hypotheses.
(b) Give the notation for the sample statistic.
(c) Give the value for the sample statistic.
(d) In the sample, which group has higher samartphone ownership: men or women?

Answer :

Answer:

a) Null hypothesis:[tex]p_{1} = p_{2}[/tex]    

Alternative hypothesis:[tex]p_{1} \neq p_{2}[/tex]  

b) [tex]z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}[/tex]   (1)  

Where [tex]\hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{688+671}{989+1012}=0.679[/tex]  

c) [tex]z=\frac{0.696-0.663}{\sqrt{0.679(1-0.679)(\frac{1}{989}+\frac{1}{1012})}}=1.58[/tex]    

d) For this case we see that [tex]\hat p_1 > \hat p_2[/tex] so then the answer for this cae would men

Step-by-step explanation:

Information given

[tex]X_{1}=688[/tex] represent the number of men with smartphone  

[tex]X_{2}=671[/tex] represent the number of women with smartphone

[tex]n_{1}=989[/tex] sample of men selected

[tex]n_{2}=1012[/tex] sample of women selected

[tex]p_{1}=\frac{688}{989}=0.696[/tex] represent the proportion of men with smartphone

[tex]p_{2}=\frac{671}{1012}=0.663[/tex] represent the proportion of women with smartphone

[tex]\hat p[/tex] represent the pooled estimate of p

z would represent the statistic

[tex]p_v[/tex] represent the value

Part a

We want to test if we have difference in the proportion owning a smartphone between men and women, the system of hypothesis would be:  

Null hypothesis:[tex]p_{1} = p_{2}[/tex]    

Alternative hypothesis:[tex]p_{1} \neq p_{2}[/tex]    

Part b

The statistic for this case is given by:

[tex]z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}[/tex]   (1)  

Where [tex]\hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{688+671}{989+1012}=0.679[/tex]  

Part c

Replacing the info given we got:

[tex]z=\frac{0.696-0.663}{\sqrt{0.679(1-0.679)(\frac{1}{989}+\frac{1}{1012})}}=1.58[/tex]    

Part d

For this case we see that [tex]\hat p_1 > \hat p_2[/tex] so then the answer for this cae would men

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