Independent Samples - In-Class Problems

1.

a) $95\%$ C.I. for $\mu_{M} - \mu_{W}$

Where

$\mu_{M}$ = mean height for men in population

$\mu_{W}$ = mean height for women in population

$s_{p} = \sqrt{\frac{(14)(1.97)^{2} + (19)(2.30)^{2}}{15 + 20 - 2}} = 2.16$

t has 15 + 20 - 2 = 33 df. Using table use t30 or t40

t*(30) = 2.042

t*(40) = 2.021

$71.2 - 64.93 \pm 2.021(2.16)\sqrt{\frac{1}{15} + \frac{1}{20}}$

$6.27 \pm 1.49$

(4.78 inches, 1.49 inches)

b)

$H_{0}: \ \mu_{M} - \mu_{W} = 4$

$H_{A}: \ \mu_{M} - \mu_{W} > 4$

$t = \frac{71.2 - 64.93 - 4}{2.16\sqrt{\frac{1}{15} + \frac{1}{20}}} = \frac{2.27}{0.7378} = 3.07$

t has 33 df, 0.001 < p < 0.0025

Very strong evidence against H0.


2.

$H_{0}: \ \mu_{S} = \mu_{N}$ or $ \mu_{S} - \mu_{N} = 0$

$H_{A}: \ \mu_{S} > \mu_{N}$ or $ \mu_{S} - \mu_{N} > 0$

Where

$\mu_{S}$ = mean number of scientists and inventors chosen by science students.

$\mu_{N}$ = mean number of scientists and inventors chosen by non-science students.

$s_{p} = \sqrt{\frac{(219)(1.571)^{2} + (96)(1.591)^{2}}{220 + 97 - 2}} = 1.577$

$t = \frac{4.614 - 4.268}{1.577\sqrt{\frac{1}{220} + \frac{1}{97}}} = \frac{0.346}{0.192} = 1.80$

t has 220 + 97 - 2 = 315 df, 0.025 < p < 0.05


3.

$99\%$ C.I. for $\mu_{I} - \mu_{II}$

Where

$\mu_{I}$ = mean breaking load for wooden beams. (Type I)

$\mu_{II}$ = mean breaking load for wooden beams. (Type II)

$s_{p} = \sqrt{\frac{14(252) + 14(258)}{28}} = \sqrt{\frac{252 + 258}{2}} = 15.97$

NOTE: if n1 = n2, $s_{p} =\sqrt{\frac{s^{2}_{1} + s^{2}_{2}}{2}}$

t has 15 + 15 - 2 = 28 df, t* = 2.763

$99\%$ C.I. for $\mu_{I} - \mu_{II}$

$1560 - 1600 \pm 2.763(15.97)\sqrt{\frac{1}{15} + \frac{1}{15}}$

$-40 \pm 16.11$

(-46.11, -23.89)

$H_{0}: \ \mu_{I} = \mu_{II}$ or $\mu_{I} - \mu_{II} = 0$

$H_{A}: \ \mu_{I} \neq \mu_{II}$ or $\mu_{I} - \mu_{II} \neq 0$

At $\alpha = 0.01$ reject H0 since 0 is not in the confidence interval for $\mu_{I} - \mu_{II}$


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Jonathan Payne
1999-04-06