The null and alternate hypotheses are: H0: ????1 ???? ????2 H1: ????1 ???? ????2 A random sample of 15 observations from the first population revealed a sample mean of 350 and a sample standard deviation of 12. A random sample of 17 observations from the second population revealed a sample mean of 342 and a sample standard devi- ation of 15. At the .10 significance level, is there a difference in the population means?

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Answer:

[tex]t=\frac{(350-342)-0}{\sqrt{\frac{12^2}{15}+\frac{15^2}{17}}}}=1.674[/tex]  

The degreess of freedom are given by:

[tex] df = 15+12-2 =25[/tex]

Now we can calculate the p value with the following probability:

[tex]p_v =2*P(t_{25}>1.674)=0.107[/tex]  

For this case since the p value is higher than the significance level we can FAIL to reject the null hypothesis and we can conclude that the true means are NOT significantly different at 10% of significance.

Step-by-step explanation:

Information provided

[tex]\bar X_{1}=350[/tex] represent the mean for sample 1  

[tex]\bar X_{2}=342[/tex] represent the mean for sample 2  

[tex]s_{1}=12[/tex] represent the sample standard deviation for 1  

[tex]s_{2}=15[/tex] represent the sample standard deviation for 2  

[tex]n_{1}=15[/tex] sample size for the group 2  

[tex]n_{2}=17[/tex] sample size for the group 2  

[tex]\alpha=0.1[/tex] Significance level provided

t would represent the statistic

Hypothesis to test

We want to verify if the true means for this case are significantly different, the system of hypothesis would be:  

Null hypothesis:[tex]\mu_{1}-\mu_{2}=0[/tex]  

Alternative hypothesis:[tex]\mu_{1} - \mu_{2}\neq 0[/tex]  

The statistic is given by:

[tex]t=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}}[/tex] (1)  

And the degrees of freedom are given by [tex]df=n_1 +n_2 -2=15+17-2=30[/tex]  

Replacing the info given we got:

[tex]t=\frac{(350-342)-0}{\sqrt{\frac{12^2}{15}+\frac{15^2}{17}}}}=1.674[/tex]  

The degreess of freedom are given by:

[tex] df = 15+12-2 =25[/tex]

Now we can calculate the p value with the following probability:

[tex]p_v =2*P(t_{25}>1.674)=0.107[/tex]  

For this case since the p value is higher than the significance level we can FAIL to reject the null hypothesis and we can conclude that the true means are NOT significantly different at 10% of significance.

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