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) (b) Test statistic for the problem above is 20-0.4286 20 = 3.03 O%0 to = 3.03 20 = -3.03

A random sample of 51 adult coyotes in a region of northern Minnesota showed the average age to be x = 2.01 years, with sample standard deviation s = 0.76 years. However, it is thought that the overall population mean age of coyotes is u = 1.75. Do the sample data indicate that coyotes in this region of northern Minnesota tend to live longer than the average of 1.75 years? Use a = 0.01. (a) What is the level of significance? State the null and alternate hypotheses. O Ho: H 1.75 yr; Hi: и 1.75 yr; Hi: и %3D 1.75 yr II Но: и 3 1.75 уr; Hi: и > 1.75 yr Ho: H = 1.75 yr; H1: µ # 1.75 yr (b) What sampling distribution will you use? Explain the rationale for your choice of sampling distribution. O The standard normal, since the sample size is large and o is known. The Student's t, since the sample size is large and o is known. The standard normal, since the sample size is large and o is unknown. O The Student's t, since the sample size is large and o is unknown. What is the value of the…

A sample of size 31 was gathered from high school seniors to estimate how many intended to attend the state university. It is known that the proportion of the population answering "yes" was 0.7. What are the mean and standard deviation, and standard error of the mean of this sample? mean = E(p) = %3D standard error = SD(p) (accurate to 3 decimal places)

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) (d) Conclusion of the test above is Reject the null hypothesis and the mean daily yield for the new process is less than that of the old procedure. Reject the null hypothesis and the mean daily yield for the new process is not less than that of the old procedure. Do not reject the null hypothesis and the mean daily yield for the new process is less than that of the old procedure. Do not reject the null hypothesis and the mean daily yield for the new process is not less than that of the old procedure.

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)?

From generation to generation, the mean age when smokers first start to smoke varies. However, the standard deviation of that age remains constant around 2.1 years. A survey of 40 smokers of this generation was done to see if the mean starting age is at least 19. The sample mean was 18.1 with a sample standard deviation of 1.3. Does the data support the claim at the 5% level? (Give the p-value, your decision, the reason for your decision and your conclusion)

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.

On a recent quiz, the class mean was 74 with a standard deviation of 2.8. Calculate the z-score (to 2 decimal places) for a person who received score of 64.

A psychologist is studying smokers' self-images, which she measures by the self-image (SI) score from a personality inventory. For adults in the U.S., the mean SI score from this inventory is about 150.  The psychologist gathers a random sample of 16 SI scores of smokers and finds that their mean is 130 and their standard deviation is 31. Assume that the population of SI scores of smokers is normally distributed with mean μ. Based on the sample, can the psychologist conclude that μ is different from 150? Use the 0.05 level of significance. Perform a two-tailed test. Then fill in the table attached. Carry your intermediate computations to at least three decimal places and round your answers as specified in the table.

The lifetime of a Michelin defender tire (in thousands of miles) is normally distributed with mean μ = 41 and standard deviation σ = 7. What proportion of Michelin tires have lifetimes between 35 and 47 thousand miles?