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2 months ago

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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 th

e 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?

1 answer:

Inessa [12.5K]2 months ago

4 0

Answer:

a) Null hypothesis: $p_{1} = p_{2}$

Alternative hypothesis: $p_{1} \neq p_{2}$

b) $z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}$ (1)

Where $\hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{688+671}{989+1012}=0.679$

c) $z=\frac{0.696-0.663}{\sqrt{0.679(1-0.679)(\frac{1}{989}+\frac{1}{1012})}}=1.58$

d) In this scenario, we notice that $\hat p_1 > \hat p_2$ thus the conclusion for this case would indicate

Step-by-step explanation:

Information provided

$X_{1}=688$ denote the number of men possessing smartphones

$X_{2}=671$ signify the number of women possessing smartphones

$n_{1}=989$ group of men sampled

$n_{2}=1012$ group of women sampled

$p_{1}=\frac{688}{989}=0.696$ symbolize the proportion of men with smartphones

$p_{2}=\frac{671}{1012}=0.663$ symbolize the proportion of women with smartphones

$\hat p$ denote the pooled estimate of p

z would denote the test statistic

$p_v$ signify the value

Part a

The objective is to evaluate if there is a disparity in smartphone ownership between men and women; the hypothesis statements would be:

Null hypothesis: $p_{1} = p_{2}$

Alternative hypothesis: $p_{1} \neq p_{2}$

Part b

The statistic relevant to this case is expressed as:

$z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}$ (1)

Where $\hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{688+671}{989+1012}=0.679$

Part c

By substituting the provided information, we find:

$z=\frac{0.696-0.663}{\sqrt{0.679(1-0.679)(\frac{1}{989}+\frac{1}{1012})}}=1.58$

Part d

In this instance, it is evident that $\hat p_1 > \hat p_2$ thus the conclusion for this case would seem

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