Question

Randomly selected students were given five seconds to estimate the value of a product of numbers with the results shown below. Estimates from students given 1×2×3×4×5×6×7×8: 50, 1000, 175, 1252, 600, 1500, 10000, 500, 40320, 800 Estimates from students given 8×7×6×5×4×3×2×1:

550, 10000, 300, 2500, 52836, 2000, 100000, 3000, 49000, 23410

Use a 0.05 significance level to test the claim that the two populations have the same mean.

(a) The test statistic is

(b) The conclusion is

A. There is not sufficient evidence to warrant the rejection of the claim that the two populations have the same mean

B. There is sufficient evidence to warrant the rejection of the claim that the two populations have the same mean

Answer 1

$\\ \bar{x_1}=5619.7, s^2_1=157332527.6\\\\ \bar{x_2}=24359.6, s^2_2=1102934708.27$

$\\\\ H_0:\mu_1=\mu_2\\\\ H_1:\mu_1\neq \mu_2$

$\\\\ Pooled~Variance(s)=\sqrt{\frac{(n1-1)S^2_1+(n2-1)S^2_2}{n1+n2-1}}=\sqrt{\frac{(10-1)*157332527.6+(10-1)*1102934708.27}{10+10-2}}\\\\ S=25102.46$

$\\\\ (a)~ test~ statistic(t)=\frac{\bar{x_1}-\bar{x_2}}{S*\sqrt{\frac{1}{n1}}+\frac{1}{n2}}=\frac{5619.7-24359.6}{25102.46*\sqrt{\frac{1}{10}}+\frac{1}{10}}=-1.67$

b.)Degree of freedom=min(n1-1, n2-1)=min(9,9)=9

P-value=P(|t|<-1.67)

=2*P(t<-1.67)

= .1293

here,

p-value > Significance(alpha=0.05), fail to reject Null hypothesis.

Ans:A. There is not sufficient evidence to warrant the rejection of the claim that the two populations have the same mean