Respuesta :
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:
[tex]H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2\neq 0[/tex]
The significance level is α=0.05.
The difference between sample means is Md=0.2.
[tex]M_d=M_1-M_2=1.15-0.95=0.2[/tex]
The estimated standard error of the difference between means is computed using the formula:
[tex]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[/tex]
Then, we can calculate the t-statistic as:
[tex]t=\dfrac{M_d-(\mu_1-\mu_2)}{s_{M_d}}=\dfrac{0.2-0}{0.069}=\dfrac{0.2}{0.069}=2.9037[/tex]
The degrees of freedom for this test are:
[tex]df=n_1+n_2-1=25+25-2=48[/tex]
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):
[tex]P-value=2\cdot P(t>2.9037)=0.0056[/tex]
As the P-value (0.0056) is smaller than the significance level (0.05), the effect is significant.
The null hypothesis is rejected.
