360 hope that this helped
Answer:
The mean and the standard deviation of the sampling distribution of the number of students who preferred to get out early are 0.533 and 0.82
Step-by-step explanation:
According to the given data we have the following:
Total sample of students= 150
80 students preferred to get out 10 minutes early
Therefore, the mean of the sampling distribution of the number of students who preferred to get out early is = 80/150 = 0.533
Therefore, standard deviation of the sampling distribution of the number of students who preferred to get out early= phat - p0/sqrt(p0(1-p)/)
= 0.533-0.5/sqrt(0.5*0.5/15))
= 0.816 = 0.82
Answer:
z = 2.1784 > 1.96,
Reject the null hypothesis
Step-by-step explanation:
For the males:
n1 = 162, x1 = 63
P1 = x 1/ n1 = 0.3889
For the Females:
n2 = 333,
x2 = 97
P 2 = x2/n2
= 0.2913
P 1= P2 Null hypothesis
P 1 is not equal to P 2 alternative hypothesis
Pooled proportion:
P= (x1 + x2) /( n1+ n2)
= (63 + 97) / (162 + 333)= 0.3232
Test statistics :
Z= (p1 - p2) /√p(1-p)× (1/n1 + 1/n2)
0.3889- 0.2913 / √0.3232 × 0.6768 × (1/162 +1/333)
=2.1784
c) Critical value :
Two tailed critical value, z critical = Norm.S .INV (0.05/2) = 1.960
Reject H o if z < -1.96 or if z > 1.96
d) Decision:
z = 2.1784 > 1.96,
Reject the null hypothesis
we have
we know that
so
the answer is in thousandths, because the denominator of the multiplication of the two fractions is one thousand
