Respuesta :
Answer:
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 [tex]n_{1} = 400 and n_{2} = 300[/tex]
Proportion of mean [tex]p_{1} = \frac{236}{400} = 0.59 and p_{2} = \frac{156}{300} = 0.52[/tex]
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
[tex]Z = \frac{p_{1} -p_{2} }{\sqrt{pq(\frac{1}{n_{1} }+\frac{1}{n_{2} )} } }[/tex]
where [tex]p = \frac{n_{1}p_{1} +n_{2}p_{2} }{n_{1}+n_{2}}= \frac{400X0.59+300X0.52}{700}[/tex]
on calculation we get p = 0.56
now q =1-p = 1-0.56=0.44
[tex]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} }[/tex]
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 = [tex]p = \frac{n_{1}p_{1} +n_{2}p_{2} }{n_{1}+n_{2}}= \frac{400X0.59+300X0.52}{700}[/tex]
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)
From a total population of 700 people who are used in this survey, the test statistics is 1.85, the critical value is 0.975 and the p value is 0.0644.
Population N = 700
n1 = females = 400
n2 = men = 300
x1: females who shop through device = 236
x2: males who shop through device = 156
p1 = x1/n1 = 236/400 = 0.59
p2 = x2/n2 = 156/300 = 0.52
The hypothesis used in the study
Null hypothesis
H0: p1 = p2
Alternate hypothesis
H1:p1 ≠ p2
test statistics calculation
[tex]\frac{0.59-0.52}{\sqrt{\frac{0.59(1-0.59)}{400} +\frac{0.52(1-0.52)}{300} } }[/tex]
= 1.85
level of significance = 5%
This is a two sample test
1-0.005/2 = 1-0.025
= 0.975
p value
2 x p(1.85)
2x(1-0.9678)
= 0.0644
The p value is greater than the level of signicance. We fail to reject the null hypothesis.
c. Find the Margin of error
z critical at 0.05 = 1.96
[tex]1.96*\sqrt{0.59*(1-0.59)/400+0.52*(1-0.52)/300}[/tex]
= 0.0743
Confidence interval
(0.59-0.52) - 0.0743,
(0.59+0.52) + 0.0743
(0.0043, 0.1443)
There is 95% confidence that the males and the females that buy clothes from the devices fall within this interval.
Read more on sample proportions here: https://brainly.com/question/870035
