Question

please find the test statistic and p value

0.2 Save Score: 0.2 of 1 pt 5 of 8 (8 complete) HW Score: 5696, 5.6 of 10 pts 8.2.15-T Assigned Media Question Help A researcher claims that the stomachs of blue crabs from Location A contain more fish than the stomachs of blue crabs from Location B. The stomach contents of a sample of 14 blue crabs from Location A contain a mean of 196 milligrams of fish and a standard deviation of 38 milligrams. The stomach contents of a sample of 10 blue crabs from Location contain a mean of 187 milligrams of fish and a standard deviation of 42 milligrams. At a=0.05, can you support the researcher's claim? Assume the population variances are equal. Complete parts (a) through (d) below H₂:14-120 HH-30 (b) Find the standardized tot statistic for 12 11.006) (Round to three decimal places as needed Enter your answer in the answer box and then click Check Answer Check Answer All 2 parts remaining

Answer 1

Since , the population varinaces are equals.

Therefore , use pooled varinace t-test.

Now , df=degrees of freedom=n1+n2-2=14+10-2=22

The pooled estimate of standard deviation is ,

$S_{p}=\sqrt{\frac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}}=\sqrt{\frac{(14-1)38^2+(10-1)42^2}{14+10-2}}=39.6851$

The null and alternative hypothesis is ,

$H_0:\mu_1-\mu_2=0$

$H_a:\mu_1-\mu_2>0$

The test is right-tailed test.

The test statistic is ,

196 – 187 tstat 0.548 X1 - X2 SpV+ 39.6851 * 11 V ito n2 10

The p-value is ,

p-value=$P(t_{df}>|t_{stat}|)=P(t_{22}>0.548)=0.2946$ ; The Excel function is , =TDIST(0.548,22,1)