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hammer [34]
3 years ago
11

Which list shows all the factors of 36

Mathematics
1 answer:
melisa1 [442]3 years ago
5 0
1,2,3,4,6,9,12,18,36
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A survey of 700 adults from a certain region​ asked, "What do you buy from your mobile​ device?" The results indicated that 59​%
Kryger [21]

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 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

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

p1 = p2

<u>Alternative hypothesis H1:</u>- 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

<u>conclusion</u>:-

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

p1 = p2

c) <u>95% confidence intervals</u>

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 <u>confidence intervals  (0.523 ,0.596)</u>

<u></u>

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