solve the problem please show step by step solutions Q 1)A variable of two populations has a mean of 8.9 and a standard deviation of 3 for one of the populations and a mean of 9.3 and a standard...

solve the problem please show step by step solutions

Q 1)A variable of two populations has a mean of 8.9 and a standard deviation

of 3 for one of the populations and a mean of 9.3 and a standard

deviation of 3.8 for the other population. For independent samples of

sizes 4 and 5, respectively, find the mean of x1 - x2.

Q 2 )Preliminary data analyses indicates that use of a paired t -test is reasonable. Perform the hypothesis test   solve the problem please show step by step solutions

by using either the critical-value approach or the P-value approach as indicated. Assume that the null

hypothesis is H0 : ?1 = ?2.

The table below shows the weights, in pounds, of seven subjects before

and after following a particular diet for two months.

Subject    A        B         C        D         E       F        G

Before    164    199     196    191     193    184      168

After      157    190    194    196    179    186      156

At the 1% significance level, do the data provide sufficient evidence to

conclude that the diet is effective in reducing weight? Use the P-value

approach.

Asked on by paki123

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

A variable of two populations has a mean of 8.9 and a standard deviation of 3 for one of the populations and a mean of 9.3 and a standard deviation of 3.8 for the other population. Independent samples of sizes 4 and 5, respectively are taken from the two populations. The mean of the two samples has to be determined.

The mean of the two samples is equal to `(8.9*4 + 9.3*5)/9 = 9.12` . The standard deviation is `sqrt(4*3^2 + 5*3.8^2) = 10.4`

Sources:
noonaa's profile pic

noonaa | (Level 1) Honors

Posted on

Answer of question 1:

Combined mean of 2 samples = (n1x1+n2x2)/n1+n2

n1 = 4, x1 = 8.9;

n2 = 5,x2 = 9.3

Combined mean =( (4×8.9)+(5×9.3))/4+5

                               = 9.12

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