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OlgaM077 [116]
3 years ago
7

We are testing H0: µ ≤ 42; versus HA: µ > 42. When Picture = 45, s = 1.2, and n = 15, at α = .01 we do not reject the null hy

pothesis. (Assume that the population from which the sample is selected is normally distributed.)
True

False
Mathematics
1 answer:
Montano1993 [528]3 years ago
5 0

Answer:

False

Step-by-step explanation:

H0: µ ≤ 42

HA: µ > 42

S represents the sample standard deviation, then we must use the t-student test.

t-statistic formula:  

t= (xbar-m)/(S/(sqrt(n)))  

xbar: sample mean

m: hypothesized value  

S: sample standard deviation  

n: number of observations  

t=(45-42)/(1.2/sqrt(15))  

t-statistic= 9.682

The critical value from the t-student distribution with, 15-1 degrees of freedom and 1% significance level , is 2.6245

Because the t-statistic is greater than the critical value, we must reject the null hypothesis.

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A professor has learned that three students in his class of 20 will cheat on the final exam. He decides to focus his attention o
balu736 [363]

Answer:

a

P(X \ge 1) = 0.509

b

P(X  \ge 1) = 0.6807

Step-by-step explanation:

From the question we are told that

   The number of students in the class is  N  =  20  (This is the population )

   The number of student that will cheat is  k =  3

   The number of students that he is focused on is  n  =  4

Generally the probability distribution that defines this question is the  Hyper geometrically distributed because four students are focused on without replacing them in the class (i.e in the generally population) and population contains exactly three student that will cheat.

Generally  probability mass function is mathematically represented as

      P(X = x) =  \frac{^{k}C_x * ^{N-k}C_{n-x}}{^{N}C_n}

Here C stands for combination , hence we will be making use of the combination functionality in our calculators  

Generally the that  he finds at least one of the students cheating when he focus his attention on four randomly chosen students during the exam is mathematically represented as

      P(X \ge 1) =  1 - P(X \le 0)

Here  

   P(X \le 0) =  \frac{ ^{3} C_0 *  ^{20 - 3} C_{4- 0}}{ ^{20}C_4}

   P(X \le 0) =  \frac{ ^{3} C_0 *  ^{17} C_{4}}{ ^{20}C_4}

   P(X \le 0) =  \frac{ 1 *  2380}{ 4845}

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    P(X \ge 1) =  1 - 0.491

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Generally the that  he finds at least one of the students cheating when he focus his attention on six randomly chosen students during the exam is mathematically represented as

    P(X \ge 1) =  1 - P(X \le 0)

   P(X  \ge 1) =1- [  \frac{^{k}C_x * ^{N-k}C_{n-x}}{^{N}C_n}]

Here n =  6

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    P(X  \ge 1) =1- [  \frac{^{3}C_0 * ^{17}C_{6}}{^{20}C_6}]

    P(X  \ge 1) =1- [  \frac{1  *  12376}{38760}]

    P(X  \ge 1) =1- 0.3193

    P(X  \ge 1) = 0.6807

   

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