Statistics 1060: Practice Solutions Week 9

: 5.67
$P(\frac{750}{12} < \overline{x} < \frac{825}{12}) = P(-1.732 < Z < 2.598) = 0.9537$
(table value: 0.9535)
: 6.1
: a) $\sigma_{\overline{x}} = \frac{4.5}{\sqrt{24}} \doteq 0.9186$ kg.
: b) $\overline{x} = 61.791\overline{6}$ , so the $95\%$ confidence interval is $\overline{x} \pm 1.96 \sigma_{\overline{x}} \doteq 59.99$ to 63.59 kg. Since 65 kg is well above the upper confidence limit, we have good evidence that $\mu < 65$ kg.
: 6.3
: a) 1 kg is 2.2 pounds, so $\overline{x}^{*} = (2.2)(61.791\overline{6}) \doteq 135.942$ lbs.
: b $\sigma_{\overline{x}^{*}} \doteq (2.2)(0.9186) \doteq 0.2021$ lbs.
: c Either compute $\overline{x}^{*} \pm 1.96 \sigma_{\overline{x}^{*}}$ , or convert the confidence limits from
6.1: 132.0 to 139.9 lbs.
: 6.7
$11.79 \pm (2.576)\left(\frac{3.2}{\sqrt{114}}\right) \doteq 11.78 \pm 0.77$ , or 11.01 to 12.55 years.
: 6.9
$35.091 \pm (1.960)\left(\frac{11}{\sqrt{44}}\right) \doteq 35.091 \pm 3.250$ , or 31.84 to 38.34.
: 6.10
$n = \left(\frac{(2.576)(3.2)}{1}\right)^{2} \doteq 67.95$ - take n = 68.
: 6.11
: a) $1.96\frac{\sigma}{\sqrt{100}} = 2.352$ points.
: b) $1.96\frac{\sigma}{\sqrt{10}} \doteq 7.438$ points.
: c) $n = \left(\frac{1.98\sigma}{3}\right)^{2} \doteq 61.46$ - take n = 62, which is under the 100 student maximum.

Jonathan Payne
1999-03-10