Mathematics, 10.01.2020 06:31 madisonwr

The net weights (in grams) of a sample of bottles filled by a machine manufactured by edne, and the net weights of a sample filled by a similar machine manufactured by orno, inc., are:

edne 8 7 6 9 7 5
orno 10 7 11 9 12 14 9 8

testing the claim at the 0.05 level that the mean weight of the bottles filled by the orno machine is greater than the mean weight of the bottles filled by the edne machine, what is the critical value? assume equal standard deviations for both samples.

a. +2.179
b. +2.145
c. +1.782
d. +1.761

Answers: 2

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Since is a right tailed test we need to look on the t distribution with 12 degrees of freedom that accumulates 0.05 of the area on the right. And we can use the following excel code:

"=T.INV(0.95,12)" and we got $t_{crit}=+1.782$

Step-by-step explanation:

When we have two independent samples from two normal distributions with equal variances we are assuming that

$\sigma^2_1 =\sigma^2_2 =\sigma^2$

And the statistic is given by this formula:

$t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$

Where t follows a t distribution with n_1+n_2 -2 degrees of freedom and the pooled variance S^2_p is given by this formula:

$\S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}$

This last one is an unbiased estimator of the common variance $\sigma^2$

The system of hypothesis on this case are:

Null hypothesis: $\mu_O \leq \mu_E$

Alternative hypothesis: $\mu_O \mu_E$

Or equivalently:

Null hypothesis: $\mu_O - \mu_E \leq 0$

Alternative hypothesis: $\mu_O -\mu_E 0$

Our notation on this case :

n_E =6 represent the sample size for group Edne

n_O =8 represent the sample size for group Orno

$\bar X_E =7$ represent the sample mean for the group Edne

$\bar X_O =10$ represent the sample mean for the group Orno

s_E=1.414 represent the sample standard deviation for group Edne

s_O=2.268 represent the sample standard deviation for group Orno

First we can begin finding the pooled variance:

$\S^2_p =\frac{(6-1)(1.414)^2 +(8 -1)(2.268)^2}{6 +8 -2}=3.834$

And the deviation would be just the square root of the variance:

S_p=1.958

And now we can calculate the statistic:

$t=\frac{(10 -7)-(0)}{1.958\sqrt{\frac{1}{6}+\frac{1}{8}}}=2.837$

Now we can calculate the degrees of freedom given by:

df=6+8-2=12

Since is a right tailed test we need to look on the t distribution with 12 degrees of freedom that accumulates 0.05 of the area on the right. And we can use the following excel code:

"=T.INV(0.95,12)" and we got $t_{crit}=+1.782$

Since our calculated value is higher than the critical value we have anough evidence to reject the null hypothesis, and on thsi case the mean for Orno is significanlty higher than the mean for Edne at 5% of significance.

Answer from: Quest

answer: a video game has three difficulty levels: easy, medium, and expert. the game's scoring works in such a way that a player has to score a fixed number of points to advance to the next level, and then the scoring starts over. the total score needed to clear all three difficulty levels is 7,500. two times the number of points needed to clear the easy level is 500 more than the number of points needed to clear the medium level. twice the number of points needed to clear the medium difficulty level is 1,500 more than the number of points needed to clear the expert level. if x, y, and z represent the easy, medium, and expert levels, respectively, and a is the coefficient matrix of the system of equations modeling this situation, identify the inverse matrix, a-1, and the solution matrix, x, of the system of equations.

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