Question

Let X and Y denote the tarsus lengths of female and
male grackles, respectively. Assume normal distributions for X and Y. Independent samples are selected from these populations and suppose

n X = 17	x ¯ = 32	s X = 7
n Y = 12	y ¯ = 34	s Y = 6

Test at 5% significance level whether the true mean tarsus length of female grackles is smaller than that of male grackles. What is the rejection region for this test?

Answer 1

The provided sample means are shown below:

Also, the provided sample standard deviations are:

and the sample sizes are

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: ​

Ha:

This corresponds to a left-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used.

(2) Rejection Region

Based on the information provided, the significance level is α=0.05, and the degrees of freedom are df=27. In fact, the degrees of freedom are computed as follows, assuming that the population variances are equal.

Hence, it is found that the critical value for this left-tailed test is tc​=−1.703, for α=0.05 and df=27.

The rejection region for this left-tailed test is R={t:t<−1.703}.

(3) Test Statistics

Since it is assumed that the population variances are equal, the t-statistic is computed as follows:

(4) Decision about the null hypothesis

Since it is observed that t=−0.802≥tc​=−1.703, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p=0.2147, and since p=0.2147≥0.05, it is concluded that the null hypothesis is not rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean μ1​ is less than μ2​, at the 0.05 significance level.

Graphically

Let me know in the comments if anything is not clear. I will reply ASAP! Please do upvote if satisfied!