7.)
a) Because this is a right-tailed test, we determine the critical value at the α=0.05 level of significance to be z sub 0.05 = 1.645. Because the test statistic is greater than the critical value, reject the null hypothesis.
b) State the value of the p-value here.Because this is a right-tailed test, the P-value is the area under the standrard normal distribution to the right of the test statistic z sub 0 = 2.31. So, P-value = P(z > 2.31) ≈ 0.0104. Because the P-Value is less than the level of significance, reject the null hypothesis. 11.)
a) Because this is a two tailed, we determine the crtical values at the α=0.05 level of significance to be -z sub 0.05/2 = -z sub 0.025 = 1.96. Because the test statistic does not lie in the critical region, do no reject the null hypothesis.
b) Because this is a two-tailed test, the P-value is the area under the standard normal distribution to the lef of -|z sub 0| = -1.49 and to the right of |z sub 0| = 1.49. So, P-valuue = 2P(Z < -1.49) ≈ 2(0.0681) = 0.1362. Because the P-value is greater than the level of significance, do not reject the null hypothesis.
P-value: P(z > 0.65) ≈ 0.2578 Since 0.65 < 2.33 and 0.2578 > 0.01, do not reject the null hypothesis. There is not sufficient evidence to conclude that more than 1.9% of Lipitor users experience flulike symptoms as a side effect.
P-Value = P(z > 0.687) = 0.2460 The P-value is greater than the significance level of 0.01, 0.687 > 0.01, so the null hypothesis fails to be rejected. Therefore, it can be concluded that there is not sufficent evidence to support the claim that the percentage is higher than 0.94.
1.)
a) 2.602
b) -1.729
c) ±2.17912.)
a) Ho: μ = 63.7 inches
H1: μ > 63.7 inches
b) From the information provided, the P-value = 0.35. If the null hypothesis that μ = 63.7 inches is true, we expect a sample mean of 63.9incheses or taller in about 35/100 samples--meaning there is a 0.35 probability of obtaining a sample mean of 63.9 inches or taller from a population whose mean = 63.7 inches. So, if we obtain 100 simple random samples of size n = 45, from a population whose mean = 63.7 inches, we could expect about 35 of the samples to result in a sample mean of 63. inches.
c) The P-value of 0.35<.10, we do not reject the null hypothesis. Therefore, the mean height of women 20 years of age or older was 63.7 inches.13.)
a) Ho: μ = 22
H1: μ > 22
b) It is given that the sample is random. The sample size, 200, is at least 30. It is reasonable to assume the sampled value are independent.
c) t sub 0 ≈ 2.176. The test statistic follows a t-distribution with 200-1=199 degrees of freedom.
Using the classical approach, the critical value is t sub 0.05 = 1.653. Since 2.176>1.653, reject the null hypothesis.
Using the P-value approach, the p-value is P(t sub 0 > 2.176) ≈ 0.0154. Since, 0.0154 < 0.05, reject the null hypothesis.
d) There is sufficient evidence to conclude that students who complete the core curriculum are ready for college-level mathematics. 14.)
a) Ho: μ = 501
H1: μ < 501
b) We know that the sample is a random sample, the sample has no outliers, the population from which the sample is drawn is normally distributed or the sample size, n, is large (n≥30), and the sampled values are independent of each other, and all are requirements to perform the test using the t-distribution.
c) The test statistic value t sub 0 = -1.379 less than the critial value -t sub.10 = -1.290, it falls in the critical region. Thus, we reject the null hypothesis. There is sufficient evidence at alpha = .10 level of significance to conclude that the results suggest that students who learn English as well as another language simultaneously score worse on the SAT critical exam.
d) Therefore, we conclude that the results suggest that students who learn English as well as another language, simultaneously, score worse on the SAT critical reading exam.