10.2

7.

  1. nPo(1-Po)=200(0.3)(1-0.3)=42>10 p=75/200=0.375 Zo=0.375-0.3/square root 0.3(1-0.3)/200=2.31 because this is a right-tailed test, we determine the critical value at the alpha=0.05 level of significance to be Z0.05=1.645. Because the test statistic is greater than the critical value,so reject the hypothesis.
  2. Because this is a right-tailed test, the P-value is the area under the standard normal distribution to the right of the test statistic, Zo=2.31. So P-value=P(z>2.31)=0.0104 And because the P-value is less than the level of significance, we reject the hypothesis.
  3. we reject the hypothesis.

9.

  1. nPo(1-Po)=150(0.55)(0.45)=37.125>10 ^p=78/150=0.52 Zo=0.52-0.55/ square root 0.55(1-0.55)/150=-0.74 Because this is a left-tailed test. we determine the critical value at the alpha=0.1 level of significance to be -Z0.1=-1.28 Because the test statistic is greater than the critical value, we do not reject the hypothesis.
  2. Because this is a left-tailed test, the P-value is the area under the standard normal distribution to the right of the test statistic, Zo=-0.74. So P-value=P(z<-0,74)=0.2296
  3. Because the P-value is greater than the level of significance, we do not reject the hypothesis.

11.

  1. nPo(1-Po)=500(0.9)(1-0.9)=45>10 P^=440/500=0.88 Test statistic= 0.88-0.9/ square root 0.9(1-0.9)/500= -1.49 Because this is a two-tailed test, we determine the critical values at the alpha=0.05 level of significance to be -Z0.05/2=-Z0.025=-1.96. Because the test statistic does not lie in the critical region, do not reject the hypothesis.
  2. Because thus us a two-tailed test, the P-value is the area under the standard normal distribution to the left of -Zo=-1.49 and to the right of the Zo=1.49 so , P-value=2P(Z<-1.49)=0.1362.
  3. Because the P-value is greater than the level of significance, do not reject the null hypothesis.

13. The P-value of 0.2743 means that if the hypothesis is true, this type of result would be expected in about 27 or 28 out of 100. The observed results are not unusual. Because the P-value is large. So we don’t reject the hypothesis.

15.

  1. 320/678=0.472

  2. Ho:p=0.5 H1:p<0.5

  3. The sample is random, Because nPo(1-Po)=678(0.5)(1-0.5)=169.5>10

  4. Skip

  5. The test statistic is Zo= 0.472-0.5/ square root 0.5(1-0.5)/678=-1.45 The P-value os P(Z<-1.45)=0.0735

  6. The P-value means that if the hypothesis is true, this type of result would be expected in anout 7 or 8 out of 100 samples.

  7. Because 0.0735> 0.05 so we do not reject the hypothesis. These is not sufficient evidence to conclude that Jim Cramer’s prediction are correct less than half of the time.

17.

  1. Hypothesis: Ho:p=0.019 H1:p>0.019 It is random selection: nPo(q-Po)=863(0.019)(1-0.019)=16.1>10 P^=19/863=0.022 Test Statistic: Zo=0.022-0.019/ square root 0.019(1-0.019)/863=0.65 Critical Value: Z0.01=2.33

P-value P(Z>0.65)=0.2578 b) since 0.65<2.33 and 0.2578>0.01, So we do not reject the hypothesis. There is not sufficient evidence to conclude that more than1.9% of Liptor users experience flulike symptoms as a side effect.

19.

  1. Hypothesis: Ho:p=0.36 H1:p >0.36 It is random selected: nPo(1-Po)=105(0.36)(1-0.36)=24.192>10 P^=51/105=0.486 Test Statistic: Zo= 0.486-036/ square root 0.36(1-0.36)/105=2.69 Critical value: Z0.05=1.645 P-value: P(Z>2.69)=0.0036

  2. Since 2.69>1.645 and 0.0036<0.05 so we reject the hypothesis. There is sufficient evidence to conclude that Hawaii has a higher proportion of traffic fatalities involviing a positive BAC than the United States as a whole.