Chapter 7. Hypothesis Testing with One Sample

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232 Chapter 7: Hypothesis Testing with One Sample2. Weights of Babiesa. This is the area to the right of 3000 under the normal curve with 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 with 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 = 1987.7 /16 = 124.231The standard deviation is s = nΣ(x 2 ) − (Σx) 2216 ⋅ 254543.43− (1987.7)= = 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 with 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 with 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 with 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
Chapter 7: Hypothesis Testing with One Sample 233d. The claim is that the population standard deviation of systolic blood pressure of women infected with thenew viral strain is 13.1, so this is a two-tailed test. The sample size is n = 16 making the degrees of freedomdf = 15. The significance level is 0.05.H 0 : σ = 13.1H 1 : ≠ σ 13.12(n −1)s 15⋅The test statistic is χ 2 22.5232= = = 44.341σ 2 13.1 2In a two-tailed test at the 0.05 significance level with df = 15, the critical values are χ 2 = 6.262 andχ 2 = 27.488.In the row for 15 degrees of freedom, the test statistic lies above 32.801, so the P-value is less than 0.01.We reject the null hypothesis.There is sufficient evidence to warrant the rejection of the claim that the population standard deviation ofsystolic blood pressure of women infected with the new viral strain is 13.1.e. Yes, the variability of the blood pressures has greatly increased.
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Chapter 7. Hypothesis Testing with One Sample

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