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Marta_Voda [28]
3 years ago
5

Suppose your statistics instructor gave six examinations during the semester. You received the following grades (percent correct

): 79, 64, 84, 82, 92, and 77. Instead of averaging the six scores, the instructor indicated he would randomly select two grades and compute the final percent correct based on the two percents. How many different samples, without replacement, of two test grades are possible
Mathematics
1 answer:
zheka24 [161]3 years ago
7 0

Answer:

15 samples

Step-by-step explanation:

The total sample space consists of 6 items

{79,64,84,82,92,77}

So,

n=6

The instructor has to randomly select 2 test scores out of 6.

So, r=6

The arrangement of scores selection doesn't matter so combinations will be used.

C(n,r)=\frac{n!}{r!(n-r)!} \\C(6,2)=\frac{6!}{2!(6-2)!}\\=\frac{6!}{2!*4!}\\=\frac{6*5*4!}{2!*4!} \\=\frac{30}{2}\\=15\ ways

Therefore, there are 15 different samples are possible without replacement ..

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nancy and lisa played 20 sets of tennis. nancy won 12 of them. write the ratio of nancy's wins to the total number of sets in si
kirill [66]

Total play = 20

Nancy win = 12

Ratio would be: 12/20 = 6/10 = 3/5

So, your answer is 3/5

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3 years ago
Identify thenull hypothesis, alternative hypothesis, test statistic, P-value or critical value(s), then state the conclusion abo
miv72 [106K]

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.

3 0
4 years ago
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