Assignment 2

ENSC 280 – Assignment 2 Solutions

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1. Peanut Butter Weight – Hypothesis Test

• Given:
  • $\bar{x}=11.92$
  • $s=0.3$
  • $n=36$
  • $\mu =12$
  • $\alpha =0.01$
• Assumptions: Normal distribution, use $z$-test

Hypotheses:

• $H_0$: $\mu =12$
• $H_1$: $\mu \ne 12$ (two-tailed)

Test Statistic:

$$z=\frac{\bar{x}-\mu }{s/\sqrt{n}}=\frac{11.92-12}{0.3/\sqrt{36}}=\frac{-0.08}{0.05}=-1.6$$

Critical Value (for $\alpha =0.01$ two-tailed):

• $z_{\alpha /2}=\pm 2.576$

Conclusion:

• $|z|=1.6<2.576\Rightarrow$ Fail to reject $H_0$
• P-value: $\approx 0.1096$
  • Since $p>0.01$, not significant

No evidence that the mean is different from 12 oz

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2. Confidence Interval for Std Dev – Height Data

• Given:
  • $n=16$
  • $s=2.4$
  • Confidence intervals for population std. dev. use chi-square

(a) 95% CI:

• df = $n-1=15$
• ${\chi }_{0.025}^2=6.262$, ${\chi }_{0.975}^2=27.488$
• Formula:

$$\left(\sqrt{\frac{(n-1)s^2}{{\chi }_{0.975}^2}},\sqrt{\frac{(n-1)s^2}{{\chi }_{0.025}^2}}\right)$$ $$=\left(\sqrt{\frac{15\cdot 2.4^2}{27.488}},\sqrt{\frac{15\cdot 2.4^2}{6.262}}\right)=(1.79,4.17)$$

(b) 99% CI:

• ${\chi }_{0.005}^2=4.601$, ${\chi }_{0.995}^2=30.578$
• Same formula → Result:

$$=\left(\sqrt{\frac{15\cdot 2.4^2}{30.578}},\sqrt{\frac{15\cdot 2.4^2}{4.601}}\right)=(1.63,4.59)$$

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3. Rope Strength Claim – t-Test

• Given:
  • $\bar{x}=7750$, $s=145$, $n=6$, $\mu =8000$, $\alpha =0.05$

Hypotheses:

• $H_0$: $\mu =8000$
• $H_1$: $\mu \ne 8000$

Test Statistic:

$$t=\frac{7750-8000}{145/\sqrt{6}}=\frac{-250}{59.21}\approx -4.22$$

Critical value ($df=5$): $\pm 2.571$

Conclusion:

• $|t|>2.571\Rightarrow$ Reject $H_0$
• Manufacturer’s claim not supported

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4. Light Bulb Supplier Comparison – z-Test

• Given:
  • $n_A=n_B=100$
  • ${\bar{x}}_A=1180$, $s_A=120$
  • ${\bar{x}}_B=1160$, $s_B=40$
  • $\alpha =0.05$

Hypotheses:

• $H_0$: ${\mu }_A={\mu }_B$
• $H_1$: ${\mu }_A\ne {\mu }_B$

Test Statistic:

$$z=\frac{1180-1160}{\sqrt{\frac{120^2}{100}+\frac{40^2}{100}}}=\frac{20}{\sqrt{144+16}}=\frac{20}{\sqrt{160}}\approx \frac{20}{12.65}\approx 1.58$$

Critical value: $\pm 1.96$

Conclusion:

• $z=1.58<1.96\Rightarrow$ Fail to reject $H_0$
• No significant difference between brands

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5. Cotton Shipment – Binomial Proportion Test

• Given:
  • Past proportion $p_0=0.4$
  • $x=20$, $n=40$ → $\hat{p}=0.5$
  • $\alpha =0.05$

Hypotheses:

• $H_0$: $p=0.4$
• $H_1$: $p>0.4$

Test Statistic:

$$z=\frac{0.5-0.4}{\sqrt{\frac{0.4\cdot 0.6}{40}}}=\frac{0.1}{\sqrt{0.006}}\approx \frac{0.1}{0.077}\approx 1.30$$

Critical value: $z=1.645$ (one-tailed)

Conclusion:

• $1.3<1.645\Rightarrow$ Fail to reject $H_0$
• Not enough evidence for quality decline

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6. Marketing Campaign Effectiveness – Proportion Difference Test

• Given:
  • Before: $p_1=0.20$, $n_1=200$
  • After: $p_2=0.27$, $n_2=300$

Hypotheses:

• $H_0$: $p_1=p_2$
• $H_1$: $p_1\ne p_2$

Pooled Proportion:

$$p=\frac{40+81}{500}=\frac{121}{500}=0.242$$

Test Statistic:

$$z=\frac{0.2-0.27}{\sqrt{0.242\cdot (1-0.242)\cdot (\frac{1}{200}+\frac{1}{300})}}=\frac{-0.07}{\sqrt{0.242\cdot 0.758\cdot 0.0083}}\approx \frac{-0.07}{0.0397}\approx -1.76$$

Critical value: $\pm 1.96$

Conclusion:

• $|z|=1.76<1.96\Rightarrow$ Fail to reject $H_0$
• Not enough evidence campaign was successful

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7. Sample Size to Detect ±0.1 lbs Error

• Given:
  • $\sigma =0.05$
  • Difference to detect: $\delta =0.1$
  • Power: 95% → $z_{\beta }=1.645$
  • Significance: $\alpha =0.05$ → $z_{\alpha }=1.645$

Sample Size Formula:

$$n={\left(\frac{(z_{\alpha }+z_{\beta })\cdot \sigma }{\delta }\right)}^2={\left(\frac{(1.645+1.645)\cdot 0.05}{0.1}\right)}^2={\left(\frac{0.1645}{0.1}\right)}^2=2.71\Rightarrow \boxed{3}$$