Question

n Assume that both populations are normally distributed. a) Test whether Hq * u2 at the a=0.10 level of significance for the given sample data. b) Construct a 90% confidence interval about M1 M2 Sample 1 19 14.5 3.2 Sample 2 19 16.5 4.7 х S

(c) Determine the test statistic (t = ?) and the P-value (P = ?). Round to two decimal places as needed.

Answer 1

a)

The null and alternate hypothesis are:

H0: $\mu_1=\mu_2$

Ha: $\mu_1\neq \mu_2$

The test statistic is given by:

$t=\frac{\overline{x_{1}}-\overline{x_{2}}}{\sqrt{\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}}} =\frac{14.5-16.5}{\sqrt{\frac{3.2^{2}}{19}+\frac{4.7^{2}}{19}}} =-1.53$

Since this is a two-tailed test, so the p-value is given by:

2 x Patnı+12-2<-1.53) = 2 x P(+19+19–2 < -1.53) = 2 x P(t36 < -1.53) = 2 x P(t36 > 1.53) = 2 x 0.067379 = 0.134758

Since p-value is greater than 0.10, so we do not have sufficient evidence to reject the null hypothesis
H0.

So, we cannot conclude $\mu_1\neq \mu_2$ .

b)

We know that the 90% confidence interval for difference in population means is given by:

$[(\overline{x_{1}}-\overline{x_{2}})\pm t_{n_{1}+n_{2}-2}(0.10/2)\sqrt{\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}}]$

$[(14.5-16.5)\pm t_{36}(0.05)\sqrt{\frac{3.2^{2}}{19}+\frac{4.7^{2}}{19}}]$

$[(14.5-16.5)\pm (1.688)\sqrt{\frac{3.2^{2}}{19}+\frac{4.7^{2}}{19}}]$

$[-2\pm 2.2019]$

c)

From part (a),

Test statisitc = -1.53

p-value = $0.134758\approx 0.13$