Gastric freezing was once a recommended treatment for ulcers in the upper intest
ID: 3315273 • Letter: G
Question
Gastric freezing was once a recommended treatment for ulcers in the upper intestine. Use of gastric freezing stopped after experiments showed it had no effect. One randomized comparative experiment found that 23 of the 77 gastric-freezing patients improved, while 25 of the 73 patients in the placebo group improved. We can test the hypothesis of "no difference" between the two groups in two ways: using the Two Sample Z Test for a difference in proportions or using the Chi-Square Test of Independence. (a) Present the data in a 2 x 2 table. Improved Gastric-freezing 23 Placebo Did not improve 54 48 25 We will start by using the Two Sample Z Test for a difference in proportions. (b) State the hypotheses (use a two-sided alternative): Ha: P1 (c) The test statistic is z = 0.17 X . (Use 2 decimal places) (d) The p-value is 0.6800 . (Use 4 decimal places)Explanation / Answer
a.
Given that,
sample one, x1 =23, n1 =77, p1= x1/n1=0.299
sample two, x2 =25, n2 =73, p2= x2/n2=0.342
null, Ho: p1 = p2
alternate, H1: p1 != p2
level of significance, = 0.05
from standard normal table, two tailed z /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 =(0.299-0.342)/sqrt((0.32*0.68(1/77+1/73))
zo =-0.574
| zo | =0.574
critical value
the value of |z | at los 0.05% is 1.96
we got |zo| =0.574 & | z | =1.96
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -0.5743 ) = 0.5658
hence value of p0.05 < 0.5658,here we do not reject Ho
ANSWERS
---------------
b.
null, Ho: p1 = p2
alternate, H1: p1 != p2
c.
test statistic: -0.574
critical value: -1.96 , 1.96
decision: do not reject Ho
d.
p-value: 0.5658
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