Question

a cognitive scientist wants to test whether there is any difference in puzzle - solving abilities between crows and sparrows. independent samples of nine crows and eight sparrows were chosen at random. the crows took a mean of 28 minutes to solve a certain puzzle, with a standard deviation of 5.2 minutes. the sparrows took a mean of 30 minutes to solve the same puzzle, with a standard deviation of 5.6 minutes. assume that the two populations of completion times are normally distributed, and that the population variances are equal. can we conclude, at the 0.05 level of significance, that μ₁, the mean puzzle - solving times for crows, is less than μ₂, the mean puzzle - solving times for sparrows? perform a one - tailed test. then complete the parts below. carry your intermediate computations to three or more decimal places and round your answers as specified in the table. (if necessary, consult a list of formulas.) (a) state the null hypothesis h₀ and the alternative hypothesis h₁. h₀:□ h₁:□ (b) determine the type of test statistic to use. select (c) find the value of the test statistic. (round to three or more decimal places.)□ (d) find the p - value. (round to three or more decimal places.)□ (e) can we conclude that the mean puzzle - solving times for crows is less than the mean puzzle - solving times for sparrows? oyes ono

Explanation:

Step1: State the hypotheses

The null hypothesis \(H_0\) is that there is no difference or the mean for crows is greater than or equal to the mean for sparrows. The alternative hypothesis \(H_1\) is that the mean for crows is less than the mean for sparrows.
\(H_0:\mu_1\geq\mu_2\)
\(H_1:\mu_1 < \mu_2\)

Step2: Determine the test - statistic type

Since the population variances are equal and the populations are normally distributed, and we have independent samples, we use a two - sample t - test.

Step3: Calculate the pooled variance \(s_p^2\)

The formula for the pooled variance is \(s_p^2=\frac{(n_1 - 1)s_1^2+(n_2 - 1)s_2^2}{n_1 + n_2-2}\), where \(n_1 = 9\), \(s_1 = 5.2\), \(n_2=8\), \(s_2 = 5.6\).
\(s_p^2=\frac{(9 - 1)\times5.2^2+(8 - 1)\times5.6^2}{9 + 8-2}=\frac{8\times27.04+7\times31.36}{15}=\frac{216.32+219.52}{15}=\frac{435.84}{15}\approx29.056\)

Step4: Calculate the t - statistic

The formula for the t - statistic is \(t=\frac{\bar{x}_1-\bar{x}_2}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}\), where \(\bar{x}_1 = 28\), \(\bar{x}_2 = 30\), \(s_p=\sqrt{29.056}\approx5.390\), \(n_1 = 9\), \(n_2 = 8\).
\(t=\frac{28 - 30}{5.390\sqrt{\frac{1}{9}+\frac{1}{8}}}=\frac{- 2}{5.390\sqrt{\frac{8 + 9}{72}}}=\frac{-2}{5.390\sqrt{\frac{17}{72}}}\approx\frac{-2}{5.390\times0.486}\approx\frac{-2}{2.620}\approx - 0.763\)

Step5: Calculate the p - value

The degrees of freedom is \(df=n_1 + n_2-2=9 + 8-2 = 15\). For a one - tailed t - test with \(t=-0.763\) and \(df = 15\), using a t - distribution table or software, the p - value is \(P(t_{15}<-0.763)\approx0.230\)

Step6: Make a decision

Since the p - value \(0.230>0.05\) (the significance level), we fail to reject the null hypothesis.

Answer:

(a) \(H_0:\mu_1\geq\mu_2\), \(H_1:\mu_1 < \mu_2\)
(b) Two - sample t - test
(c) \(-0.763\)
(d) \(0.230\)
(e) No