Mathematics, 30.11.2019 02:31, shealwaysknows23

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.

Answers: 3

Show answers

Answers

With the p value obtained and using the significance level given $\alpha=0.01$ we have $p_v\alpha$ 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

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

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 $\simga^2$

The system of hypothesis on this case are:

Null hypothesis: $\mu_1 = \mu_2$

Alternative hypothesis: $\mu_1 \neq \mu_2$

Or equivalently:

Null hypothesis: $\mu_1 - \mu_2 = 0$

Alternative hypothesis: $\mu_1 -\mu_2 \neq 0$

Our notation on this case :

n_1 =20 represent the sample size for group 1

n_2 =24 represent the sample size for group 2

$\bar X_1 =8.91$ represent the sample mean for the group 1

$\bar X_2 =8.23$ represent the sample mean for the group 2

s_1=2.1 represent the sample standard deviation for group 1

s_2=2.5 represent the sample standard deviation for group 2

First we can begin finding the pooled variance:

$\S^2_p =\frac{(20-1)(2.1)^2 +(24 -1)(2.5)^2}{20 +24 -2}=5.418$

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

S_p=2.328

2) Calculate the statistic

And now we can calculate the statistic:

$t=\frac{(8.91-8.23)-(0)}{2.328\sqrt{\frac{1}{20}}+\frac{1}{24}}=0.965$

Now we can calculate the degrees of freedom given by:

df=20+24-2=42

3) Calculate the p value

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

$p_v =2*P(t_{42}0.965) =0.340$

4) Conclusion

So with the p value obtained and using the significance level given $\alpha=0.01$ we have $p_v\alpha$ 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.

answered: Guest

answer: the value should be used for i in the formula of calculation the monthly payment , the interest rate per period is 0.85.

step-by-step explanation:

the given rate of interest i= 85%

in calculating the monthly payment for a loan with an interest rate compounded monthly, the value of the interest rate per period we use in the formula must be in decimal form.

to convert it in decimal, we divide it by 100, we get

$i=\frac{85}{100}=0.85$