Question

In: Statistics and Probability

A survey of 800 adults from a certain region​ asked, "What do you buy from your...

A survey of 800 adults from a certain region​ asked, "What do you buy from your mobile​ device?" The results indicated that 57​% of the females and 48​% of the males answered clothes. The sample sizes of males and females were not provided. Suppose that of 500 ​females, 285 reported they buy clothing from their mobile​ device, while of 300 ​males, 144 reported they buy clothing from their mobile device.

D: What are your answers to​ (a) through​ (c) if 432 males said they buy clothing from their mobile​ device?

Determine the value of the test statistic Determine the value of X2=

Determine the proportion of items of interest in sample​ 2, p2=

ZSTAT=

The critical values for this test of hypothesis =

​p-value=

Construct and interpret a90%, 95%, and 99​% confidence interval estimate for the difference between the proportion of males and females who said they buy clothing from their mobile device.

Solutions

Expert Solution

Answer:

for the given data,

a)

the hypotheses are

Ho: p1 - p2 = 0         

Ha: p1 - p2 ╪ 0                           

For female:

sample size,     n1=   500         

no. of successes, sample 1 =     x1=   285         

proportion success of sample 1 , p̂1=   x1/n1=   0.5700                           

For male:

sample size,     n2 =    300         

no. of successes, sample 2 =     x2 =    144         

proportion success of sample 1 , p̂ 2=   x2/n2 =    0.4800                           

difference in sample proportions:

p̂1 - p̂2 = 0.57 - 0.48 = 0.09   

pooled proportion , p =   (x1+x2)/(n1+n2)=   0.5363      

std error ,SE = SQRT(p*(1-p)*(1/n1+ 1/n2) =   0.0364        

Z-statistic = (p̂1 - p̂2)/SE = (0.09/ 0.0364) =   2.47

z-critical value :

Z 0.05 = 1.96 (Using Z-table)

b)

for this, p-value = 0.0135 (Using calculator)

As p-value < 0.05, we reject H0.

Therefore, there is sufficient evidence to conclude that there is difference between males and females in the proportion who said they buy clothing from their mobile device

c)

i)

for 90% level, level of significance, α =   0.10             

Z critical value =   Z α/2 =    1.645 (Using z table)                 

Std error , SE = SQRT(p̂1 * (1 - p̂1)/n1 + p̂2 * (1-p̂2)/n2) =     0.0364         

margin of error , E = Z*SE = 1.645   *   0.0364   =   0.0598                 

90% confidence interval :           

lower limit = (p̂1 - p̂2) - E =    0.090 -   0.0598   =   0.0302 = 0.03

upper limit = (p̂1 - p̂2) + E =    0.090   +   0.0598   =   0.1498 = 0.15   

ii)

for 95% level, level of significance, α =   0.05              

Z critical value =   Z α/2 =    1.960   (Using z table)     

margin of error , E = Z*SE =    1.960   *   0.0364   =   0.0713                 

95% confidence interval:                  

lower limit = (p̂1 - p̂2) - E =    0.090   -   0.0713   =   0.019

upper limit = (p̂1 - p̂2) + E =    0.090   +   0.0713   =   0.161

iii)

for 99% level, level of significance, α =   0.01             

Z critical value =   Z α/2 =    2.576   [(Using z table)                   

margin of error , E = Z*SE =    2.576   *   0.0364   =   0.0937                 

99% confidence interval:                  

lower limit = (p̂1 - p̂2) - E =    0.090 -   0.0937   =   -0.004

upper limit = (p̂1 - p̂2) + E =    0.090   +   0.0937   =   0.184

                 


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