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Burka [1]
3 years ago
12

The grade appeal process at a university requires that a jury be structured by selecting five individuals randomly from a pool o

f eight students and ten faculty.
What is the probability of selecting a jury of all students?
Mathematics
1 answer:
valkas [14]3 years ago
8 0

Answer:

0.006535 (to 6 decimal place)

Step-by-step explanation:

There are

8 students in total

10 faculty in total

Jury consists of 5 members

We need to select jury of students

Hence, the probability would be:

ways of selecting 5 students from 8 students DIVIDED by ways of selecting 5 members from total of 18

So,

5 of 8 selection = 8C5

5 out of 18 selection = 18C5

So,

Probability of selecting a jury of all students = 8C5/18C5 = 56/8568 = 0.006535

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Test the claim that the proportion of people who own cats is significantly different than 90% at the 0.1 significance level.
VikaD [51]

Answer:

1. H0 : p = 0.9

   H1 : p ≠ 0.9

2. The test is two tailed.

3. Reject the null hypothesis

Step-by-step explanation:

We are given that we have to test the claim that the proportion of people who own cats is significantly different than 90% at the 0.1 significance level.

So, Null Hypothesis, H_0 : p = 0.90

Alternate Hypothesis, H_1 : p \neq 0.9

Here, the test is two tailed because we have given that to test  the claim that the proportion of people who own cats is significantly different than 90% which means it can be less than 0.90 or more than 0.90.

Now, test statistics is given by;

            \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } } ~ N(0,1)   , where,  n = sample size = 100

                                                       \hat p = 0.94 (given)

So, Test statistics = \frac{0.94-0.90}{\sqrt{\frac{0.94(1-0.94)}{100} } } = 1.68

Now, P-value = P(Z > 1.68) = 1 - P(Z <= 1.68)

                                           = 1 - 0.95352 = 0.0465 ≈ 0.05 or 5%

Now, our decision rule is that;

       If p-value < significance level  ⇒ Reject null hypothesis

       If p-value > significance level  ⇒ Accept null hypothesis

Since, here p-value is less than significance level as 0.05 < 0.1, so we have sufficient evidence to reject null hypothesis.

Therefore, we conclude that proportion of people who own cats is significantly different than 90%.

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