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3a) $\bar{x} = 15.4375$ $s_0 = 0.2973$ 3b) $80\% = 0.8$ $\alpha = 1 - 0.8 = 0.2 \to P(|Z| \ge \alpha) = P(|Z| \ge 0.2) = 0.8$ $\frac{\alpha}{2} = 0.1 \to P(|Z| \ge 0.1) = 0.4$ Table 2 $\to Z = 1.28$ $P(-Z, \le Z \le Z_0.1) = 0.8$ $P(-1.28 \le \frac{15.4375 - \mu}{0.2973/\sqrt{8}} \le 1.28) = 0.8$ $-1.28 \le \frac{15.4375 - \mu}{0.2973/\sqrt{8}} \le 1.28 (\frac{0.2973}{\sqrt{8}})$ $-0.1345 \le 15.4375 - \mu \le 0.1345$ $0.1345 \le 15.4375 - \mu \le 0.1395$ $15.572 \le \mu \le 15.303$ 3c) t-test comparison confident interval $\bar{x}_1 = 15.4375$ $\bar{x}_2 = 14.6$ $\alpha = 0.2$ $s_1 = 0.2975$ $s_2 = 0.24$ $\frac{\alpha}{2} = 0.1$ $n_1 = 8$ $n_2 = 10$ $t_{cal} = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} = 2.6479 \approx$ $JF = F = \frac{s_1^2/s_2^2}{(n_1 - 1)/(n_2 - 1)} = 13.324 = 13$ Table 5: $t_{table} = 1.350$ $\therefore t_{cal} > t_{table} \to statistically \ different$

Name: 3a barx 154375 s0 02973 3b 80 08 alpha 1 08 02 to pz ge alpha pz ge 02 08 fracalpha2 01 to pz ge 01 04 table 2 to z 128 p z le z le z01 08 p 128 le frac154375 mu02973sqrt8 le 128 08 128 le 66257
Uploaded: 2023-01-31T16:39:48-08:00
Channel: Rashmi Sinha
Description: 3a barx 154375 s0 02973 3b 80 08 alpha 1 08 02 to pz ge alpha pz ge 02 08 fracalpha2 01 to pz ge 01 04 table 2 to z 128 p z le z le z01 08 p 128 le frac154375 mu02973sqrt8 le 128 08 128 le 66257

          3a) $\bar{x} = 15.4375$
$s_0 = 0.2973$
3b) $80\% = 0.8$
$\alpha = 1 - 0.8 = 0.2 \to P(|Z| \ge \alpha) = P(|Z| \ge 0.2) = 0.8$
$\frac{\alpha}{2} = 0.1 \to P(|Z| \ge 0.1) = 0.4$
Table 2 $\to Z = 1.28$
$P(-Z, \le Z \le Z_0.1) = 0.8$
$P(-1.28 \le \frac{15.4375 - \mu}{0.2973/\sqrt{8}} \le 1.28) = 0.8$
$-1.28 \le \frac{15.4375 - \mu}{0.2973/\sqrt{8}} \le 1.28 (\frac{0.2973}{\sqrt{8}})$
$-0.1345 \le 15.4375 - \mu \le 0.1345$
$0.1345 \le 15.4375 - \mu \le 0.1395$
$15.572 \le \mu \le 15.303$
3c) t-test comparison
confident interval
$\bar{x}_1 = 15.4375$
$\bar{x}_2 = 14.6$
$\alpha = 0.2$
$s_1 = 0.2975$
$s_2 = 0.24$
$\frac{\alpha}{2} = 0.1$
$n_1 = 8$
$n_2 = 10$
$t_{cal} = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} = 2.6479 \approx$
$JF = F = \frac{s_1^2/s_2^2}{(n_1 - 1)/(n_2 - 1)} = 13.324 = 13$
Table 5: $t_{table} = 1.350$
$\therefore t_{cal} > t_{table} \to statistically \ different$

$3a barx 154375 s0 02973 3b 80 08 alpha 1 08 02 to pz ge alpha pz ge 02 08 fracalpha2 01 to pz ge 01 04 table 2 to z 128 p z le z le z01 08 p 128 le frac154375 mu02973sqrt8 le 128 08 128 le 66257$

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