Mathematics, 28.03.2020 03:43 sethlynn2003

A survey of 700 adults from a certain region asked, "What do you buy from your mobile device?" The results indicated that 59% of the females and 52% of the males answered clothes. The sample sizes of males and females were not provided. Suppose that of 400females, 236 reported they buy clothing from their mobile device, while of 300 males, 156 reported they buy clothing from their mobile device. Complete parts (a) through (d) below. A)-Determine the value of the test statistic- Determine the critical values (two tailed)b) Find the p valuec) construct a 95% confidence interval estimate for the difference between the proportion of males and females who said they buy clothing from their mobile devices.

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Answer from: bamagirl3871

a) the test statistic z = 1.891

the null hypothesis accepted at 95% level of significance

b) the critical values of 95% level of significance is zα =1.96

c) 95% of confidence intervals are (0.523 ,0.596)

Step-by-step explanation:

A survey of 700 adults from a certain region

Given sample sizes $n_{1} = 400 and n_{2} = 300$

Proportion of mean $p_{1} = \frac{236}{400} = 0.59 and p_{2} = \frac{156}{300} = 0.52$

Null hypothesis H0 : assume that there is no significant difference between males and women reported they buy clothing from their mobile device

p1 = p2

Alternative hypothesis H1:- p1 ≠ p2

a) The test statistic is

$Z = \frac{p_{1} -p_{2} }{\sqrt{pq(\frac{1}{n_{1} }+\frac{1}{n_{2} )} } }$

where $p = \frac{n_{1}p_{1} +n_{2}p_{2} }{n_{1}+n_{2}}= \frac{400X0.59+300X0.52}{700}$

on calculation we get p = 0.56

now q =1-p = 1-0.56=0.44

$Z = \frac{p_{1} -p_{2} }{\sqrt{pq(\frac{1}{n_{1} }+\frac{1}{n_{2} )} } }\\ =\frac{0.56-0.52}{\sqrt{0.56X0.44}(\frac{1}{400}+\frac{1}{300} }$

after calculation we get z = 1.891

b) The critical value at 95% confidence interval zα = 1.96 (from z-table)

The calculated z- value < the tabulated value

therefore the null hypothesis accepted

conclusion:-

assume that there is no significant difference between males and women reported they buy clothing from their mobile device

p1 = p2

c) 95% confidence intervals

The confidence intervals are P± 1.96(√PQ/n)

we know that = $p = \frac{n_{1}p_{1} +n_{2}p_{2} }{n_{1}+n_{2}}= \frac{400X0.59+300X0.52}{700}$

after calculation we get P = 0.56 and Q =1-P =0.44

Confidence intervals are ( P- 1.96(√PQ/n), P+ 1.96(√PQ/n))

now substitute values , we get

( 0.56- 1.96(√0.56X0.44/700), 0.56+ 1.96(0.56X0.44/700))

on simplification we get (0.523 ,0.596)

Therefore the population proportion (0.56) lies in between the 95% of confidence intervals (0.523 ,0.596)

Answer from: Quest

35 is an outlier

step-by-step explanation:

Answer from: Quest

step-by-step explanation: the declaration of the rights of man and of the citizen (french: déclaration des droits de l'homme et du citoyen de 1789), set by france's national constituent assembly in 1789, is a human civil rights document from the french revolution.[1]

the declaration was drafted by the abbé sieyès and the marquis de lafayette, in consultation with thomas jefferson.[2] influenced by the doctrine of "natural right", the rights of man are held to be universal: valid at all times and in every place, pertaining to human nature itself. it became the basis for a nation of free individuals protected equally by the law. it is included in the beginning of the constitutions of both the fourth french republic (1946) and fifth republic (1958) and is still current. inspired by the enlightenment philosophers, the declaration was a core statement of the values of the french revolution and had a major impact on the development of freedom and democracy in europe and worldwide.[3]

the 1789 declaration, together with the 1215 magna carta, the 1689 english bill of rights (1689), the 1776 united states declaration of independence, and the 1789 united states bill of rights, inspired in large part the 1948 united nations universal declaration of human rights