Question

Part 2: Open Ended 11.17 points] The following sample data are based on a report by the Pew Research Organization in early 2018. A random sample of 1250 men included 313 who say they are online "almost constantly." Another random sample of 1100 women said the same. Test the claim at 5% significance that the proportions of women and men who are online "almost constantly" are different Step 1: Give the null and alternative hypotheses Describe pi= Describe p= H. H. level of significance Right Tail OR Left Tail OR Two Tail? Step 2: Collect the data Is the data from a random sample? Are the criteria for normality satisfied? n, = P: = n2 = Which is larger? p, OR P - P= p2 Step 3: Assess the evidence Test statistic Z= P-value p= Plot the test statistic and shade the area that represents the P-value. Step 4: State the conclusion. •Interpret the P-value and compare it to the level of significance. Decide whether the data tells us that we can reject or fail to reject the null hypothesis. Using complete sentences, write a conclusion explaining what you have learned about whether the sample data supports the claim.

Answer 1

given that
sample one, x1 =313, n1 =1250, p1= x1/n1=0.25
sample two, x2 =313, n2 =1100, p2= x2/n2=0.285
finding a p^ value for proportion p^=(x1 + x2 ) / (n1+n2)
p^=0.266
q^ Value For Proportion= 1-p^=0.734
null, Ho: p1 = p2
alternate, H1: p1 != p2
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic (z) = (p1-p2)/√(p^q^(1/n1+1/n2))
zo =(0.25-0.285)/sqrt((0.266*0.734(1/1250+1/1100))
zo =-1.868
| zo | =1.868
critical value
the value of |z α| at los 0.05% is 1.96
we got |zo| =1.868 & | z α | =1.96
make decision
hence value of |zo | < | z α | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -1.8684 ) = 0.0617
hence value of p0.05 < 0.0617,here we do not reject Ho
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[ANSWERS]
p1= proportion of men who are constantly online
p2= proportion of women who are constantly online
two tailed test
step2: x1 =313, n1 =1250, p1= x1/n1=0.25
x2 =313, n2 =1100, p2= x2/n2=0.285
test statistic: -1.868
critical value: -1.96 , 1.96
decision: do not reject Ho
p-value: 0.0617, we don't sopport the claim

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