Suppose that the following are a random sample of length-of-life measurements of components whose length of life has an exponential distribution with mean 0. 1105, 4612, 75, 684, 83, 965, 364, 1011, 197, 309 The reliability of a component at time t is F(t) = e-t/. Use the large sample approximation (even though n is only 10) for the distribution of the MLE for to compute a 97% confidence interval for F(1100). |< F(1100) <
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- 26 The table to the right shows the cost per ounce (in dollars) for a random sample of toothpastes exhibiting very good stain removal, good stain removal, and fair stain removal. At α=0.01, can you conclude that the mean costs per ounce are different? Perform a one-way ANOVA test by completing parts a through d. Assume that each sample is drawn from a normal population, that the samples are independent of each other, and that the populations have the same variances. Very good stain removal Good stain removal Fair stain removal 0.37 0.75 0.60 0.49 2.66 1.18 0.33 0.46 0.46 1.64 0.33 0.50 0.58 0.41 1.39 (b) Identify the degrees of freedom for the numerator and for the denominator, determine the critical value, and determine the rejection region. The degrees of freedom for the numerator, d.f.N, is ____ and the degrees of freedom for the denominator, d.f.D, is _____ The critical…Suppose the lengths of the pregnancies of a certain animal are approximately normally distributed with mean u = 260 days and standard deviation o = 22 days. Complete parts (a) through (f) below. O B. If 100 pregnant individuals were selected independently from this population, we would expect 37 pregnancies to last less than 253 days. O C. If 100 pregnant individuals were selected independently from this population, we would expect pregnancies to last exactly 253 days. (b) Suppose a random sample of 21 pregnancies is obtained. Describe the sampling distribution of the sample mean length of pregnancies. The sampling distribution of x is normal with H; = 260 and o: = 4.8008. (Round to four decimal places as needed.) (c) What is the probability that a random sample of 21 pregnancies has a mean gestation period of 253 days or less? The probability that the mean of a random sample of 21 pregnancies is less than 253 days is approximately 0.0724. (Round to four decimal places as needed.)…A food manufacturer claims that eating its new cereal as part of a daily diet lowers total blood cholesterol levels. The table shows the total blood cholesterol levels (in milligrams per deciliter of blood) of seven patients before eating the cereal and after one year of eating the cereal as part of their diets. Use technology to test the mean difference. Assume the samples are random and dependent, and the population is normally distributed. At α=0.05, can you conclude that the new cereal lowers total blood cholesterol levels? Patient 1 2 3 4 5 6 7 Total Blood Cholesterol (Before) 215 225 235 240 255 260 225 Total Blood Cholesterol (After) 214 222 240 237 254 257 222 Let the blood cholesterol level before eating the cereal be population 1. Let the blood cholesterol level after eating the cereal be population 2. Identify the null and alternative hypotheses, where μd=μ1−μ2. Choose the correct…
- Suppose in a local Kindergarten through 12th grade (K -12) school district, 49% of the population favor a charter school for grades K through 5. A simple random sample of 144 is surveyed. a. Find the mean and the standard deviation of X of B(144, 0.49). Round off to 4 decimal places. O = b. Now approximate X of B(144, 0.49) using the normal approximation with the random variable Y and the table. Round off to 4 decimal places. Y - N( c. Find the probability that at most 81 favor a charter school using the normal approximation and the table. (Round off to z-values up to 2 decimal places.) P(X 75) - P(Y > a (Z > e. Find the probability that exactly 81 favor a charter school using the normal approximation and the table. (Round off to z-values up to 2 decimal places.) P(X = 81) - P(Time to Complete the Course Right 45 47 50 49 50 50 48 44 Left | 45 | 43 48 47 50 52 | 44 42 Assume a Normal distribution. What can be concluded at the the a = 0.01 level of significance level of significance? For this study, we should use t-test for the difference between two dependent population means a. The null and alternative hypotheses would be: p1 p2 Ні: p1 p2 b. The test statistic t v v = 1.199 (please show your answer to 3 decimal places.) c. The p-value = [1338 your answer to 4 decimal places.) d. The p-value is > va e. Based on this, we should fail to reject f. Thus, the final conclusion is that ... * (Please show v the null hypothesis. O The results are statistically insignificant at a = 0.01, so there is insufficient evidence to conclude that the population mean time to complete the obstacle course with a patch overSuppose in a local Kindergarten through 12th grade (K -12) school district, 49% of the population favor a charter school for grades K through 5. A simple random sample of 144 is surveyed. a. Find the mean and the standard deviation of X of B(144, 0.49). Round off to 4 decimal places. O = b. Now approximate X of B(144, 0.49) using the normal approximation with the random variable Y and the table. Round off to 4 decimal places. Y - N( c. Find the probability that at most 81 favor a charter school using the normal approximation and the table. (Round off to z-values up to 2 decimal places.) P(X 75) - P(Y > a (Z > e. Find the probability that exactly 81 favor a charter school using the normal approximation and the table. (Round off to z-values up to 2 decimal places.) P(X = 81) - P(Recommended textbooks for youMATLAB: An Introduction with ApplicationsStatisticsISBN:9781119256830Author:Amos GilatPublisher:John Wiley & Sons Inc
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