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3 years ago

Find the value of x such that the data set has the mean of 103. 100,119,105,111,110, x

Mathematics

1 answer:

LenKa [72]3 years ago

7 0

Answer:

105

Step-by-step explanation:

add up the terms including x it gives 545+x/6=103

multiply both sides by 6 it will gives you 545+x=618

then collect like terms you have x=618-546

then you have 73 as your answer

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4 years ago

An article reported on the results of an experiment in which half of the individuals in a group of 60 postmenopausal overweight

slavikrds [6]

Answer:

Step-by-step explanation:

From the given information:

Let's first compute the null and alternative hypothesis

Null hypothesis:

$\mathbf{H_o: \mu_1-\mu_2=1}$

Alternative hypothesis:

$H_a :\mu_1 -\mu_2 > 1$

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i.e

$n_1=n_2 = 30$

the sample mean for sample 1 $\overline x_1$ = 5.7

the standard deviation for sample 1 $s_1$ = 3.1

the sample mean for sample 2 $\overline x_2$ = 3.9

the standard deviation for sample 2 $s_2$ = 2.7

degree of freedom for this test can be computed by using the formula:

$df = \dfrac{\begin {pmatrix} \dfrac{s_1^2}{n_1} + \dfrac{s^2_2}{n_2} \end {pmatrix}^2 } {\dfrac{ (\dfrac{s_1^2}{n_1}^2)}{ n_1-1} + \dfrac{ (\dfrac{s_2^2}{n_2}^2)}{ n_2-1} }$

$df = \dfrac{\begin {pmatrix} \dfrac{3.1^2}{30} + \dfrac{2.7^2}{30} \end {pmatrix}^2 } {\dfrac{ \begin {pmatrix} \dfrac{3.1^2}{30} \end {pmatrix}^2}{ 30-1} + \dfrac{ \begin {pmatrix} \dfrac{2.7^2}{30} \end {pmatrix}^2}{ 30-1} }}$

df = 114.68

The test statistics can be computed as follows:

$t = \dfrac{ \overline x_1 -\overline x_2- (\mu_1 -\mu_2)}{\sqrt{ \dfrac{s_1^2}{n_1} +\dfrac{s_2^2}{n_2} }}$

$t = \dfrac{ 5.7-3.9- (1)}{\sqrt{ \dfrac{3.1^2}{30} +\dfrac{2.7^2}{30} }}$

t = 1.07

Using the level of significance of 0.1, the P-value for the test statistics at the df of 114.68 is:

P-value = 0.143

Decision rule: To reject the null hypothesis if the level of significance is greater than the p-value.

Conclusion: We fail to reject the null hypothesis because the p-value is greater than the level of significance at 0.1.

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