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n order to test the null hypothesis of equal means (alternative hypothesis says they are unequal), two simple random samples are obtained.In sample 1, the mean and standard deviation are 10 and 6 respectively.In sample 2, the mean and standard deviation are 16 and 4 respectively.Each sample is of size 6.The data are normally distributed.The true standard deviations of the two populations may be assumed to be equal.What can you say about the P-value for the test of equal means? --> It must be

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In order to test the null hypothesis of equal means (alternative hypothesis says they are unequal), two simple random samples are obtained.In sample 1, the mean and standard deviation are 10 and 6 respectively.In sample 2, the mean and standard deviation are 16 and 4 respectively.Each sample is of size 6.The data are normally distributed.The true standard deviations of the two populations may be assumed to be equal.What can you say about the P-value for the test of equal means? -->

It must be that the resulting P-value must be

A) Greater than 0.2

B) Between 0.19 and 0.10

C) Between 0.09 and .05

D) Between .049 and .02

E) Less than 0.02

Answer:

The correct option is B

Explanation:

From the question we are told that

The mean of sample 1 is
\= x_1 = 10

The standard deviation is
s_1 = 6

The mean of sample two is
\= x_2 = 16

The standard deviation is
s_2= 4

The sample size for both samples is
n_1 = n_2 = n = 6

The null hypothesis is
H_o : \mu_1 = \mu_2

The alternative hypothesis is
H_a : \mu_1 \\e \mu_2

Generally the test statistics is mathematically represented as


z = \frac{ \= x_1 - \= x_2 }{ \sqrt{ (s^2_1)/(n_1) +(s^2_1)/(n_2) } }

=>
z = \frac{ 10 - 16 }{ \sqrt{ (6^2)/(6) +(4^2)/(6) } }

=>
z = -1.3646

From the z table the area under the normal curve to the left corresponding to -1.3646 is


P(Z < -1.3646) = 0.086189

Generally the p-value is mathematically represented as


p-value = 2 P(Z < -1.3646)

=>
p-value = 0.086189 * 2

=>
p-value = 0.1724

So the p-value is Between 0.19 and 0.10

User Farhood ET
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