What is mine times nine

Assume that you have a sample of n 1 equals 6, with the sample mean Upper X overbar 1 equals 50, and a sample standard deviati

tigry1 [53]

Answer:

$t=\frac{(50 -38)-(0)}{7.46\sqrt{\frac{1}{6}+\frac{1}{5}}}=2.656$

$df=6+5-2=9$

$p_v =P(t_{9}>2.656) =0.0131$

Since the p value is higher than the significance level given of 0.01 we don't have enough evidence to conclude that the true mean for group 1 is significantly higher thn the true mean for the group 2.

Step-by-step explanation:

Data given

$n_1 =6$ represent the sample size for group 1

$n_2 =5$ represent the sample size for group 2

$\bar X_1 =50$ represent the sample mean for the group 1

$\bar X_2 =38$ represent the sample mean for the group 2

$s_1=7$ represent the sample standard deviation for group 1

$s_2=8$ represent the sample standard deviation for group 2

System of hypothesis

The system of hypothesis on this case are:

Null hypothesis: $\mu_1 \leq \mu_2$

Alternative hypothesis: $\mu_1 > \mu_2$

We are assuming that the population variances for each group are the same

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

The statistic for this case is given by:

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

The pooled variance is:

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

We can find the pooled variance:

$S^2_p =\frac{(6-1)(7)^2 +(5 -1)(8)^2}{6 +5 -2}=55.67$

And the pooled deviation is:

$S_p=7.46$

The statistic is given by:

$t=\frac{(50 -38)-(0)}{7.46\sqrt{\frac{1}{6}+\frac{1}{5}}}=2.656$

The degrees of freedom are given by:

$df=6+5-2=9$

The p value is given by:

$p_v =P(t_{9}>2.656) =0.0131$

Since the p value is higher than the significance level given of 0.01 we don't have enough evidence to conclude that the true mean for group 1 is significantly higher thn the true mean for the group 2.

4 0

3 years ago

3n - n + 7 = 25<br><br>Help plz

WARRIOR [948]

The answer is N=6 but first you have to simplify the problem try that '

8 0

3 years ago

Normal Distribution. Cherry trees in a certain orchard have heights that are normally distributed with mu = 112 inches and sigma

Lubov Fominskaja [6]

Answer:

The probability that a randomly chosen tree is greater than 140 inches is 0.0228.

Step-by-step explanation: