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Given the following information about a hypothesis test of the difference between two means based on independent random samples, which one of the following is the correct rejection region at a significance level of .05? Assume that the samples are obtained from normally distributed populations having equal variances.HA: μA > μB, = 12, = 9, s1 = 5, s2 = 3, n1 = 13, n2 = 10.A. Reject H0 if Z > 1.96B. Reject H0 if Z > 1.645C. Reject H0 if t > 2.08D. Reject H0 if t > 1.782E. Reject H0 if t > 1.721

Respuesta :

Answer:

Null hypothesis:[tex]\mu_{A} \leq \mu_{B}[/tex]

Alternative hypothesis:[tex]\mu_{A} > \mu_{B}[/tex]

Since we dpn't know the population deviations for each group, for this case is better apply a t test to compare means, and the statistic is given by:

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

Now we need to find the degrees of freedom given by:

[tex] df = n_A + n_B -2= 13+10-2=21[/tex]

And now since we are conducting a right tailed test we are looking ofr a value who accumulates 0.05 of the are on the right tail fo the t distribution with df =21 and we got:

[tex] t_{cric}= 1.721[/tex]

And for this case the rejection zone would be:

E. Reject H0 if t > 1.721

Step-by-step explanation:

Data given and notation

[tex]\bar X_{A}=12[/tex] represent the mean for 1

[tex]\bar X_{B}=9[/tex] represent the mean for 2

[tex]s_{A}=5[/tex] represent the sample standard deviation for 1

[tex]s_{2}=3[/tex] represent the sample standard deviation for 2

[tex]n_{1}=13[/tex] sample size for the group 1

[tex]n_{2}=10[/tex] sample size for the group 2

t would represent the statistic (variable of interest)

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

Develop the null and alternative hypotheses for this study

We need to conduct a hypothesis in order to check if the mean for group A is higher than the mean for B:

Null hypothesis:[tex]\mu_{A} \leq \mu_{B}[/tex]

Alternative hypothesis:[tex]\mu_{A} > \mu_{B}[/tex]

Since we dpn't know the population deviations for each group, for this case is better apply a t test to compare means, and the statistic is given by:

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

Now we need to find the degrees of freedom given by:

[tex] df = n_A + n_B -2= 13+10-2=21[/tex]

And now since we are conducting a right tailed test we are looking ofr a value who accumulates 0.05 of the are on the right tail fo the t distribution with df =21 and we got:

[tex] t_{cric}= 1.721[/tex]

And for this case the rejection zone would be:

E. Reject H0 if t > 1.721

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