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.

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

Proportion of mean

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

where  

on calculation we get   p = 0.56    

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

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 =

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)

it is 570

step-by-step explanation: