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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 Baseline : (mu 1 minus mu 2 )not equals 0Ha: μ1−μ2≠0. Give the significance level of the​ test, and interpret the result. Use alphaαequals=0.05. What is the test​ statistic?

User MarianD
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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.

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\\eq 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{(\sigma_1^2+\sigma_2^2)/(n)}=\sqrt{(0.31^2+0.15^2)/(25)}\\\\\\s_(M_d)=\sqrt{(0.119)/(25)}=√(0.005)=0.069

Then, we can calculate the t-statistic as:


t=(M_d-(\mu_1-\mu_2))/(s_(M_d))=(0.2-0)/(0.069)=(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.

User Jeanne Boyarsky
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