Question

for following questions please 1. compute test value 2. find critical value. thanks so much kind sirs! much apprecation!! thanks for help

Next Previous up Question 1 of 5 (1 point) View problem in a pop 9.2 Section Exercise 1a,b,c,d,e (critical value, table) Is there a significant difference at a = 0.05 in the mean heights in feet of waterfalls in Europe and the ones in Asia? The data are shown. Source:World Almanac. Use the critical value method with tables. Europe Asia 900 1246 1385 984 614 722 320 722 964 830 1312 Download data Use y, for the mean height of waterfalls in Europe. Assume the variables are normally distributed and the variances are unequal. Part 1 State the hypotheses and identify the claim with the correct hypothesis. H: not claim H: A claim This hypothesis test is a two-tailed test. Part 2 out of 5 Find the critical value(s). Enter the values in ascending order and round the answers to three i e hp

Answer 1

1)

The null and alternate hypothesis are:

H0:

μΕ E =HA

Ha: $\mu_E\neq \mu_A$

From the data, we have,

ΣτΕ TE Π 5827 5 1165.4, nE Στέ – η ΕΤΕ SE 6970341 – 5(1165.432 5 – 1 211.8698 ΠΕ - 1 ΣΤΑ 4172 6 - 695.3333, nA SA = Στ2 - nAT nA-1 3140160 - 6(695.3333)2 6 -1 = 218.7370

The test statistic is given by:

$t=\frac{\overline{x_{E}}-\overline{x_{A}}}{\sqrt{\frac{s_{E}^{2}}{n_{E}}+ \frac{s_{A}^{2}}{n_{A}}}} =\frac{1165.4-695.3333}{\sqrt{\frac{211.8698^{2}}{5}+\frac{218.7370^{2}}{6}}}=3.61$

Since this is a two-tailed test, so the critical values are given by:

$\\-t_{n_{E}+n_{A}-2}(\alpha/2)=-t_{5+6-2}(0.05/2)=-t_{9}(0.025)=-2.262\\ \\t_{n_{E}+n_{A}-2}(\alpha/2)=t_{5+6-2}(0.05/2)=t_{9}(0.025)=2.262$

Since the test statistic value does not lie between the critical values, so we have sufficient evidence to reject H0.

Thus we can conclude that $\mu_E\neq \mu_A$ .