What is 5y+=33u this is not an answer
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Answer:
a) test statistic = 2.12
b) p-value = 0.017
c) we reject the Null hypothesis
Step-by-step explanation:
Given data :
N = 200
girls (x) = 115 , Boys = 85
p = x / n = 115 / 200 = 0.575
significance level ( ∝ ) = 0.1
<em>aim : test whether the proportion of girls births after the treatment is greater than 50% that occurs without any treatment </em>.
<u>A) Determine the test statistic </u>
H0 : p = 0.5
Ha : p > 0.5
to determine the test statistic we will apply the z distribution at ( ∝ ) = 0.1
Z - test statistic = ( 0.575 - 0.5) / = 2.12
<u>b) determine the p-value</u>
The P-value can be determined using the normal standard table
P-value = 1 - p(Z< 2.12 ) = 1 - 0.9830 = 0.017
c) Given that the p value ( 0.017 ) < significance level ( 0.1 )
we will reject the H0 because there is evidence showing that proportion of girls birth is > 50%