Chapter 7. Hypothesis Testing with One Sample
Chapter 7. Hypothesis Testing with One Sample
Chapter 7. Hypothesis Testing with One Sample
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232 <strong>Chapter</strong> 7: <strong>Hypothesis</strong> <strong>Testing</strong> <strong>with</strong> <strong>One</strong> <strong>Sample</strong>2. Weights of Babiesa. This is the area to the right of 3000 under the normal curve <strong>with</strong> mean µ = 3420 and standard deviation σ =495. In order to use table A-2, we must convert into the standard normal distribution,z = x − µ 3000 − 3420= =−0.85.σ 495The probability is the area to the right of -0.85, and so the probability is P = 1 – 0.1977 = 0.8023.b. Since the babies are independent of one another, the probability that all five have weights greater than 3000g is P = (0.8023) 5 = 0.3324 .c. We use the central limit theorem to convert to the standard normal distribution. The probability will be thearea to the right of 3000 under the normal curve <strong>with</strong> mean µ x= µ = 3420 and standard deviationσ x= σ / n = 495/ 5 = 221.371.z = x − µ 3000 − 3420x= =−1.90.σ x221.371The probability is the area to the right of -1.90, and so the probability is P = 1 – 0.0287 = 0.9713.d. Looking for the area 0.90 in the body of Table A-2, we find the z score to be 1.28. We findP 90= µ + z ⋅σ = 3420 +1.28 ⋅ 495 = 4053.6.3. Blood Pressure Readingsa. The mean is x = Σx /n = 198<strong>7.</strong>7 /16 = 124.231The standard deviation is s = nΣ(x 2 ) − (Σx) 2216 ⋅ 254543.43− (198<strong>7.</strong>7)= = 22.523n(n −1)16 ⋅15b. The sample is a random sample. We will assume that the sample is from a normally distributed population.The claim is that the mean systolic blood pressure for women infected <strong>with</strong> the new viral strain is 114.8, sothis is a two-tailed test. The sample size is n = 16 making the degrees of freedom df = 15. The significancelevel is 0.05.H 0 : µ = 114.8H 1 : µ 114.8 ≠The test statistic is t = x − µs / n = 124.231−114.8 = 1.67522.523/ 16In a two-tailed test at the 0.05 significance level <strong>with</strong> df = 15, the critical values are ±t α=±t .025=±2.131.In the row for 15 degrees of freedom, the test statistic is between 1.341 and 1.753, so the P-value isbetween 0.10 and 0.20.We fail to reject the null hypothesis.There is not sufficient evidence to warrant the rejection of the claim that the mean systolic blood pressurefor women infected <strong>with</strong> the new viral strain is 114.8.c. We again assume that the data is from a normally distributed population. For a 95% confidence interval,α=.05, and so α/2=.025. Since n=16, the degrees of freedom df=n–1=15. Using Table A-3, we see thatt .025= 2.131.The margin of error is E = t .025The confidence interval isx − E < µ < x + Esn222.523= 2.131 = 11.99916124.231−11.999 < µ < 124.231+11.999112.232 < µ < 136.23The confidence interval limits do contain the value 114.8