A dance class consists of 22 students, of which 10 are women and 12 are men. if 5 men and 5 women are to be chosen and then pair
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Answer:
see below
Step-by-step explanation:
We have elected to use a table, so that the rows and columns have labels that identify the table content. A matrix can be made from the data in the table.
Answer:
Step-by-step explanation:
This is a test of 2 population proportions. Let m and w be the subscript for men and women players. The population proportions would be pm and pw
Pm - Pw = difference in the proportion of male and female players.
The null hypothesis is
H0 : pm = pw
pm - pw = 0
The alternative hypothesis is
Ha : pm ≠ pw
pm - pw ≠ 0
it is a two-tailed test
Sample proportion = x/n
Where
x represents number of success
n represents number of samples
For men,
xm = 1027
nm = 2441
Pm = 1027/2441 = 0.42
For women,
xw = 509
nw = 1273
Pw = 509/1273 = 0.4
The pooled proportion, pc is
pc = (xm + xw)/(nm + nw)
pc = (1027 + 509)/(2441 + 1273) = 0.41
1 - pc = 1 - 0.41 = 0.59
z = (Pm - Pw)/√pc(1 - pc)(1/nm + 1/nw)
z = (0.42 - 0.4)/√(0.41)(0.59)(1/2441 + 1/1273) = 0.02/√0.0002891223
z = 1.18
Since it is a 2 tailed test, we would find the p value by doubling the area to the right of the z score to include the area to left.
Area to the right from the normal distribution table is
1 - 0.881 = 0.119
P value = 0.119 × 2 = 0.238
Since 0.05 < 0.238, we would accept the null hypothesis
Therefore, there is no sufficient evidence to conclude that that men and women have unequal success in challenging calls.
