Pooled‐variances t‐test for mean comparison we have two normal populations with unknown means μ₁ and μ₂ and a common variance σ². two independent samples are randomly drawn from the populations with the following data: n₁ = 20 xത₁ = 8.91 s₁ = 2.1n₂ = 24 xത₂ = 8.23 s₂ = 2.5(a)[5] at the level of significance α = 0.01, test h₀: μ₁ = μ₂ versus h₁: μ₁ ≠ μ₂. sketch the test.

With the p value obtained and using the significance level given we have so we can conclude that we have enough evidence to FAIL reject the null hypothesis, and we can said that at 1% of significance the mean of the group 1 is not significantly different from than the mean for the group 2.

Step-by-step explanation:

1) Notation and hypothesis

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

And the statistic is given by this formula:

Where t follows a t distribution with degrees of freedom and the pooled variance is given by this formula:

This last one is an unbiased estimator of the common variance

The system of hypothesis on this case are:

Null hypothesis:

Alternative hypothesis:

Or equivalently:

Null hypothesis:

Alternative hypothesis:

Our notation on this case :

represent the sample size for group 1

represent the sample size for group 2

represent the sample mean for the group 1

represent the sample mean for the group 2

represent the sample standard deviation for group 1

represent the sample standard deviation for group 2

First we can begin finding the pooled variance:

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

2) Calculate the statistic

And now we can calculate the statistic:

Now we can calculate the degrees of freedom given by:

3) Calculate the p value

And now we can calculate the p value using the altenative hypothesis:

4) Conclusion

So with the p value obtained and using the significance level given we have so we can conclude that we have enough evidence to FAIL reject the null hypothesis, and we can said that at 1% of significance the mean of the group 1 is not significantly different from than the mean for the group 2.  

9.55 degrees.

step-by-step explanation:

the given angle is

1 cycle is equal to 2π radians. so 4π will be equal to two cycles.

5.166667π-4π = 1.166667π

hence, this angle goes a little more than negative x axis.

1.166667π-π= 0.166667π

hence, the reference angle is 0.166667π radians.

1 radian = 57.2958 degrees

now, in degrees, this should be equals to 9.55 degrees.

-10i^2-4i

step-by-step explanation:

evaluate for p=2,q=5i

(2−4)(2+5i)i

(2−4)(2+5i)i

=−10i2−4i