Suppose you want to test the claim that mu 1μ1less than

So if the difference between [tex]\bar X_{1}-\bar X_{2} <-1.072[/tex] we have enough evidence to reject the null hypothesis in our case [tex]\bar X_{1}-\bar X_{2} =-2.55<-1.072[/tex].

Step-by-step explanation:

Suppose you want to test the claim that μ1<μ2. Two samples are randomly selected from each population. The sample statistics are given below. At a level of significance of α=0.05, when should you reject Upper H0? n 1=35, n 2=42, x overbar 1=21.28, x overbar 2=23.83, σ1=2.9, sigma σ2=2.8

Data given and notation

[tex]\bar X_{1}=21.28[/tex] represent the mean for sample 1

[tex]\bar X_{2}=23.83[/tex] represent the mean for sample 2

[tex]\sigma_{1}=2.9[/tex] represent the population standard deviation for 1

[tex]\sigma_{2}=2.8[/tex] represent the population standard deviation for 2

[tex]n_{1}=35[/tex] sample size for the group 1

[tex]n_{2}=42[/tex] sample size for the group 2

[tex]\alpha=0.05[/tex] Significance level provided

z would represent the statistic (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to check if the mean for group 1 is less than the mean fro group 2, the system of hypothesis would be:

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

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

We have the population standard deviation's, and the sample sizes are large enough, so we can apply a z test to compare means, and the statistic is given by:

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

z-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]z=\frac{(21.28-23.83)-0}{\sqrt{\frac{2.9^2}{35}+\frac{2.8^2}{42}}}}=-3.902[/tex]

Critical value

We need to find a critical value on the normal standard distribution that accumulates 0.05 of the area on the left tail. And on this case we got:

[tex]z_{crit}=-1.64[/tex]

Since our calculated value [tex]|z_{calc}| >|z_{crit}|[/tex] we have enough evidence to reject the null hypothesis.

And if we are interested on the rejection zone in terms fo the difference of means we can do this:

[tex]-1.64=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}}[/tex]

And if we solve for [tex]\bar X_{1}-\bar X_{2}[/tex] we got:

[tex]\bar X_{1}-\bar X_{2} = -1.64*\sqrt{\frac{2.9^2}{35}+\frac{2.8^2}{42}}}=-1.072[/tex]

So if the difference between [tex]\bar X_{1}-\bar X_{2} <-1.072[/tex] we have enough evidence to reject the null hypothesis in our case [tex]\bar X_{1}-\bar X_{2} =-2.55<-1.072[/tex].