CHAPTER 6

HYPOTHESIS TESTS

6.2.

(35 points: 5 for each part) For the normal approximation:

• Scientific Hypothesis:Clinton approval > 0.5.
• Statistical Hypotheses:

  H₀: p = 0.5

  H_a: p > 0.5,

  p = population proportion approving of Clinton.

• Model: b(10000,p).
• Standardized Test Statistic:

  $$z^\ast = \displaystyle\frac{\hat{p} - 0.5}{\sqrt{(0.5)(1-0.5)/10000}} = \displaystyle\frac{0.51-0.5}{0.005} = 2,$$

• Assumptions:$np, n(1 - p) \geq 10.$ Satisfied since $n\hat{p} = 5500,\;\;\;n(1 - \hat{p}) = 4500$.
• p-value:$P(Z \geq 2) = 0.0228, \;\;\; Z \sim N(0,1)$
• Conclusions:Reject H₀ in favor of H_a.

For the exact test, change the following:

• Test Statistic: y^*=5100
• Assumptions: none needed
• p-value:$P(b(10000,0.5) \geq 5100)=0.0233$.

6.6.

The one-sided test is appropriate since the complaints were that red came up too often. If the complaints were only that the wheel wasn't fair, then a two-sided alternative hypotheses would be appropriate.

6.8.

• Scientific Hypothesis: Mean breaking strengths < 42,300
• Statistical Hypothesis:

  $$\mu$$ H₀: = 42,300

  $$\mu$$ H_a: < 42,300,

• Standardized Test Statistic:

  $$t^\ast =\displaystyle\frac{\bar{y} - 42,300}{s/\sqrt{5}} = \displaystyle\frac{42166-42300}{259.41/\sqrt{5}} = -1.1551$$

• Assumptions:Normality.
• p-value: $P(t_{4} \leq -1.1551)=0.1562$.
• Conclusions:Do not reject H₀ in favor of H_a.

6.14.

(35 points: 5 for each part)

$$\bar{y}$$ Group s n

Treatment 40.7 10.8 50

Control 39.0 12.2 50

• Scientific Hypothesis:Refrigerated batteries last longer, on average.
• Model:Two population C+E
• Statistical Hypothesis:

  $$\mu_T \mu_C$$ H₀: =

  $$\mu_T \mu_C$$ H_a: >

  $\mu_T =$ mean life of refrigerated batteries

  $\mu_C =$ mean life of unrefrigerated batteries

• Standardized Test Statistic: Since the sample sizes are large, we compute

  $$z^\ast = \frac{\bar{y}_T - \bar{y}_C} {\sqrt{\frac{s^2_T}{n_... ...0}{\sqrt{\frac{(10.8)^2}{50} + \frac{(12.2)^2}{50}}} = 0.74$$

• Assumptions: No extreme outliers.
• p-value: $P(Z \geq 0.74) = 0.2297$
• Conclusion:Do not reject H₀.

6.20.

(10 points) No. A large p-value signifies that the observed data are consistent with the null hypothesis, but it does not prove H₀ to be true. Remember, in calculating the p-value, we assume H₀ is true.

6.24.

• Scientific Hypothesis: The proportion of female students who prefer cookie A to cookie B is greater than the proportion of male students who prefer cookie A to cookie B.
• Model: The observed number of male students who prefer cookie A to cookie B is $Y_1\sim b(179,p_1)$, and the observed number of female students who prefer cookie A to cookie B is $Y_2\sim b(66,p_2)$. The populations are independent.
• Statistical Hypothesis:

  H₀: p₁ = p₂

  H_a: p₁ < p₂.

• Standardized Test Statistic: The point estimator of p₁ is $\hat{p}_1=94/179=0.53$, and of p₂ is $\hat{p}_2=39/66=0.59$. The standardized test statistic is

  where

  $$\hat{p} = \frac{y_1 + y_2}{n_1 + n_2} = \frac{94+39}{179+66} = \frac{133}{245} = 0.54.$$

• Assumptions: Since

  we are comfortable using the large sample test based on the normal approximation.

• p-value: $P(Z \leq -0.84) = 0.2005$.
• Conclusion: Since 0.2005 > 0.05, do not reject H₀. There is not enough evidence to conclude that the proportion of female students who prefer cookie A is greater than the proportion of male students who prefer cookie A.