1. In the past, a chemical company produced 880 pounds of a certain type of plastic per day. Now, using a newly developed and less expensive process, the mean daily yield of plastic for the first 50 days of production is 871 pounds; the standard deviation is 21 pounds. Do the data provide sufficient evidence to indicate that the mean daily yield for the new process is less than that of the old procedure? (Use α=0.05) (a) Alternative hypothesis for the problem above is Ομ < 871 Ομ > 871 Ομ < 880 Ομ > 880

A score of X=75 is measured in a population with a mean of μμ=100. A z-score of z=+1.50 is calculated. Without knowing the standard deviation, explain why the z-score of z=+1.50 is incorrect.

The level of calcium in the blood of healthy young adults follows a normal distribution with mean u = 10 milligrams per deciliter and standard deviation s = 0.4.  A clinic measures the blood calcium of 100 healthy pregnant young women at their first visit for prenatal care.  The sample mean of these 100 measurements is  X = 9.8.  Is this evidence that the mean calcium level in the population from which these women come is less than 10?  To answer this question, we perform the following hypothesis test: H0: u = 10, Ha: u < 10. What is the test statistic? (use two decimal places in your answer) What is the p-value equal to? At the 10% significance level, do you accept or reject the null hypothesis? Answer ACCEPT or REJECT t the 5% significance level, do you accept or reject the null hypothesis? Answer ACCEPT or REJECT

The nicotine content in cigarettes of a certain brand is normally distributed with mean (in milligrams) u and standard deviation o = 0.1. The brand advertises that the mean nicotine content of their cigarettes is 1.5 mg. Now, suppose a reporter wants to test whether the mean nicotine content is actually higher than advertised. He takes measurements from a SRS of 15 cigarettes of this brand. The sample yields an average of 1.6 mg of nicotine. Conduct a test using a significance level of a = 0.05. (a) The standardized test statistic (b) The critical value (endpoint of Rejection Region) is z* = (c) The final conclusion is OA. The nicotine content is probably higher than advertised. B. There is not sufficient evidence to show that the ad is misleading.

A sample of n = 16 scores is selected from a population with μ = 80 with σ = 20. On average, how much error would be expected between the sample mean and the population mean?​

In the past, students in a particular course have achieved a mean Xmas test score of 72.6, with a standard deviation of 12.5. This year, 15 students in one section achieved a mean Xmas test score of 72.2, while 10 students in another section achieved a mean Xmas test score of 77.3. Is the overall (grand) mean of the 25 students in this year's two sections significantly different from the mean obtained in previous years (α ≤ .05)?

A company that makes cola drinks states that the mean caffeine content per​ 12-ounce bottle of cola is 40 milligrams. You want to test this claim. During your​ tests, you find that a random sample of thirty​ 12-ounce bottles of cola has a mean caffeine content of 38.9 milligrams. Assume the population is normally distributed and the population standard deviation is 7.3 milligrams. At α=0.09​, can you reject the​ company's claim? Complete parts​ (a) through​ (e).       ​(a) Identify H0 and Ha. Choose the correct answer below.     A. H0​: μ=40 Ha​: μ≠40   B. H0​: μ=38.9 Ha​: μ≠38.9   C. H0​: μ≠40 Ha​: μ=40   D. H0​: μ≤40 Ha​: μ>40   E. H0​: μ≤38.9 Ha​: μ>38.9   F. H0​: μ≠38.9 Ha​: μ=

You are conducting a study to see if the proportion of women over 40 who regularly have mammograms is significantly less than 0.66. Thus you are performing a left-tailed test. Your sample data produce the test statistic z=−1.812z=-1.812.   Find the p-value accurate to 4 decimal places.

A random sample of 17 adult male wolves from the Canadian Northwest Territories gave an average weight x1 = 98.6 pounds with estimated sample standard deviation s1 = 5.6 pounds. Another sample of 26 adult male wolves from Alaska gave an average weight x2 = 88.2 pounds with estimated sample standard deviation s2 = 6.0 pounds.   (a) Categorize the problem below according to parameter being estimated, proportion p, mean μ, difference of means μ1 – μ2, or difference of proportions p1 – p2. Then solve the problem. μp    μ1 – μ2p1 – p2 (b) Let μ1 represent the population mean weight of adult male wolves from the Northwest Territories, and let μ2 represent the population mean weight of adult male wolves from Alaska. Find a 99% confidence interval for μ1 – μ2. (Use 1 decimal place.) lower limit   upper limit   (c) Examine the confidence interval and explain what it means in the context of this problem. Does the interval consist of numbers that are all positive? all negative? of…