Question

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

Answer 1

Complete Question

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