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3 years ago

HELP ME!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Mathematics

1 answer:

Alik [6]3 years ago

8 0

The answer is B hoped it helped!! :)

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9x=180
X=20
So the answer is x =20

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3 years ago

Test the null hypothesis Upper H 0 : (mu 1 minus mu 2 )equals 0H0: μ1−μ2=0 versus the alternative hypothesis Upper H Subscript a

Law Incorporation [45]

Answer:

The test statistic t is t=2.9037.

The null hypothesis is rejected.

For a significance level of 0.05, there is enough evidence to support the alternative hypothesis.

Step-by-step explanation:

The question is incomplete:

The sample 1, of size n1=25 has a mean of 1.15 and a standard deviation of 0.31.

The sample 2, of size n2=25 has a mean of 0.95 and a standard deviation of 0.15.

This is a hypothesis test for the difference between populations means.

The null and alternative hypothesis are:

$H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2\neq 0$

The significance level is α=0.05.

The difference between sample means is Md=0.2.

$M_d=M_1-M_2=1.15-0.95=0.2$

The estimated standard error of the difference between means is computed using the formula:

$s_{M_d}=\sqrt{\dfrac{\sigma_1^2+\sigma_2^2}{n}}=\sqrt{\dfrac{0.31^2+0.15^2}{25}}\\\\\\s_{M_d}=\sqrt{\dfrac{0.119}{25}}=\sqrt{0.005}=0.069$

Then, we can calculate the t-statistic as:

$t=\dfrac{M_d-(\mu_1-\mu_2)}{s_{M_d}}=\dfrac{0.2-0}{0.069}=\dfrac{0.2}{0.069}=2.9037$

The degrees of freedom for this test are:

$df=n_1+n_2-1=25+25-2=48$

This test is a two-tailed test, with 48 degrees of freedom and t=2.9037, so the P-value for this test is calculated as (using a t-table):

$P-value=2\cdot P(t>2.9037)=0.0056$

As the P-value (0.0056) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

8 0

3 years ago