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padilas [110]
2 years ago
7

Test the claim that the mean GPA of Orange Coast students is smaller than the mean GPA of Coastline students at the 0.005 signif

icance level.
(a)The null and alternative hypothesis would be:_______
A. H 0 : μ O ≤ μ C
H 1 : μ O > μ C
B. H 0 : p O ≤ p C
H 1 : p O > p C
C. H 0 : p O ≥ p C
H 1 : p O < p C
D. H 0 : μ O = μ C
H 1 : μ O ≠ μ C
E. H 0 : p O = p C
H 1 : p O ≠ p C
F. H 0 : μ O ≥ μ C
H 1 : μ O < μ C
(b)The test is:_______
A. left-tailed
B. two-tailed
C. right-tailed
(c)The sample consisted of 60 Orange Coast students, with a sample mean GPA of 2.91 and a standard deviation of 0.05, and 60 Coastline students, with a sample mean GPA of 2.96 and a standard deviation of 0.03.
(d)The test statistic is:________ (to 2 decimals)
(e)The p-value is:________ (to 2 decimals)
(f)Based on this we:_______
A. Fail to reject the null hypothesis
B. Reject the null hypothesis
Mathematics
1 answer:
algol [13]2 years ago
5 0

Answer:

a) F. H 0 : μ O ≥ μ C

H 1 : μ O < μ C

b) B. two-tailed

d) t=\frac{(2.91-2.96)-0}{\sqrt{\frac{0.05^2}{60}+\frac{0.03^2}{60}}}}=-6.64  

e) p_v =P(t_{118}  

f) B. Reject the null hypothesis

Step-by-step explanation:

Information provided

\bar X_{O}=2.91 represent the mean for the Orange Coast

\bar X_{C}=2.96 represent the mean for the Coastline

s_{O}=0.05 represent the sample standard deviation for Orange Coast

s_{C}=0.03 represent the sample standard deviation for Coastline

n_{O}=60 sample size for Orange Coast

n_{C}=60 sample size for Coastline

\alpha=0.005 Significance level provided

t would represent the statistic

Part a

For this case we want to test the claim that the mean GPA of Orange Coast students is smaller than the mean GPA of Coastline students

Null hypothesis:\mu_{O} \geq \mu_{C}  

Alternative hypothesis:\mu_{O} < \mu_{C}  

F. H 0 : μ O ≥ μ C

H 1 : μ O < μ C

Part b

For this case we need to conduct a left tailed test.

B. two-tailed

Part d

The statistic is given by:

t=\frac{(\bar X_{O}-\bar X_{C})-\Delta}{\sqrt{\frac{s^2_{O}}{n_{O}}+\frac{s^2_{C}}{n_{C}}}} (1)  

And the degrees of freedom are given by df=n_1 +n_2 -2=60+60-2=118  

Replacing the info we got:

t=\frac{(2.91-2.96)-0}{\sqrt{\frac{0.05^2}{60}+\frac{0.03^2}{60}}}}=-6.64  

Part e

We can calculate the p value with this probability:

p_v =P(t_{118}  

Part f

Since the p value is a very low value compared to the significance level given of 0.005 we can reject the null hypothesis.

B. Reject the null hypothesis

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Answer:

no it is false.

Step-by-step explanation:

Because The left side 12 does not equal to the right side 16, which means that the given statement is false.

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3 years ago
According to a study done by a university​ student, the probability a randomly selected individual will not cover his or her mou
VikaD [51]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

 P(X = 8) =  0.0037

b

 P(X <  5) =  0.805

c

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I would be surprised because the value is very small , less the 0.05

Step-by-step explanation:

From the question we are told that

The probability a randomly selected individual will not cover his or her mouth when sneezing is p = 0.267

Generally data collected from this study follows  binomial  distribution because the number of trials is  finite , there are only two outcomes, (covering  , and  not covering mouth when sneezing ) , the trial are independent

Hence for a randomly selected variable  X we have that  

   X \ \ \~ \ \ { B ( p , n )}

The probability distribution function for binomial  distribution is  

    P(X = x ) =  ^nC_x *  p^x *  (1 -p) ^{n-x}

Considering question a

Generally the  the probability that among 12 randomly observed individuals exactly 8 do not cover their mouth when​ sneezing is mathematically represented as

     P(X = 8) =  ^{12} C_8 *  (0.267)^8 *  (1- 0.267)^{12-8}

Here C denotes  combination

So

     P(X = 8) =  495  *  0.000025828 * 0.28867947

    P(X = 8) =  0.0037

Considering question b

Generally the probability that among 12 randomly observed individuals fewer than 5 do not cover their mouth when​ sneezing is mathematically represented as

     P(X <  5 ) =[P(X = 0 ) + \cdots + P(X = 4)]

=>   P(X <  5 ) =[ ^{12} C_0 *  (0.267)^0 *  (1- 0.267)^{12-0} + \cdots +  ^{12} C_4 *  (0.267)^4 *  (1- 0.267)^{12-4} ]

=> P(X <  5 )  =  0.02406 +  0.10516 + 0.21067 + 0.25580 + 0.20964

=>  P(X <  5) =  0.805

Considering question c

Generally the probability that fewer than half(6) covered their mouth when​ sneezing(i.e the probability the greater than half do not cover their mouth when sneezing) is mathematically represented as

      P(X > 6) =  1 - p(X \le  6)

=>    P(X > 6) = 1 - [P(X = 0) + \cdots + P(X =6)]

=>    P(X > 6)=1 - [^{12} C_0 *  (0.267)^0 *  (1- 0.267)^{12-0}+ \cdots + ^{12} C_4 *  (0.267)^6 *  (1- 0.267)^{12-6} ]

=>    P(X > 6)= 1 - [0.02406 + \cdots + 0.0519 ]  

=>    P(X > 6) =  0.0206

I would be surprised because the value is very small , less the 0.05

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3 years ago
Solve the problem.
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Answer:

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Step-by-step explanation:

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Substitute:

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