MATH SOLVE

6 months ago

Q:
# A random sample of 49 measurements from one population had a sample mean of 15, with sample standard deviation 5. An independent random sample of 64 measurements from a second population had a sample mean of 18, with sample standard deviation 6. Test the claim that the population means are different. Use level of significance 0.01

Accepted Solution

A:

Answer:Comparing the p value with the significance level [tex]\alpha=0.01[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and the difference between the two groups is significantly different. Step-by-step explanation:[tex]\bar X_{1}=15[/tex] represent the mean for sample 1[tex]\bar X_{2}=18[/tex] represent the mean for sample 2[tex]s_{1}=5[/tex] represent the sample standard deviation for 1 [tex]s_{f}=6[/tex] represent the sample standard deviation for 2 [tex]n_{1}=49[/tex] sample size for the group 2 [tex]n_{2}=64[/tex] sample size for the group 2 [tex]\alpha=0.01[/tex] Significance level providedt would represent the statistic (variable of interest) Concepts and formulas to use We need to conduct a hypothesis in order to check if the difference in the population means, 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] We don't have the population standard deviation's, so for this case is better apply a t test to compare means, and the statistic is given by: [tex]t=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}}[/tex] (1)And the degrees of freedom are given by [tex]df=n_1 +n_2 -2=49+64-2=111[/tex] t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.With the info given we can replace in formula (1) like this: [tex]t=\frac{(15-18)-0}{\sqrt{\frac{5^2}{49}+\frac{6^2}{64}}}}=-2.897[/tex]P valueSince is a bilateral test the p value would be: [tex]p_v =2*P(t_{111}<-2.897)=0.0045[/tex] Comparing the p value with the significance level [tex]\alpha=0.01[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and the difference between the two groups is significantly different.