1. A machine produces metal pieces that are cylindrical in shape. A sample of 36 metal pieces is taken and the mean diameters for this sample was 6.21 centimeters. In addition from previous studies it is know that the population is approximately Normal with a standard deviation 1.84centimeter.

Note: Do the calculations by hand, but show the R code and output for the z-value, t-value or for finding area under the standard normal curve.

Note: Round your numbers to exactly 2 decimal places.

1. Construct and interpret a 95% confidence interval for the true mean diameter of these metal pieces.
2. Construct a 99% confidence interval for the true mean diameter of these metal pieces and explain the different in size between the intervals found in (a) and (b)
3. What do you think would happen to the confidence interval found in (b) if we were to decrease the sample size? Explain.

Karen wants to advertise how many chocolate chips are in each Big Chip cookie at her bakery. She randomly selects a sample of 48 cookies and finds that the number of chocolate chips per cookie in the sample has a mean of 15.2 and a standard deviation of 1.3. What is the 80% confidence interval for the number of chocolate chips per cookie for Big Chip cookies?

Enter your answers accurate to one decimal place (because the sample statistics are reported accurate to one decimal place).

A certain flight arrives on time 83 percent of the time. Suppose 179 flights are randomly selected. Use the normal approximation to the binomial to approximate the probability that

​(a) exactly 143 flights are on time.

​(b) at least 143 flights are on time.

​(c) fewer than 152 flights are on time.

​(d) between 152 and 159​， inclusive are on time.

Question 1: Test the following hypothesis: H0:μ = 41.8 versus H1:μ > 41.8; y =43.1, n =16, σ = 3.1, α =0.05. Assume that the data comes from a normal distribution. The conclusion is to

- reject the null hypothesis

- fail to reject the null hypothesis

Question 2: (not for forum discussion) Test the following hypothesis: H0:μ = 41.8 versus H1:μ ≠ 41.8; y =43.1, n =16, σ = 3.1, α =0.05. Assume that the data comes from a normal distribution. The conclusion is to

- reject the null hypothesis

- fail to reject the null hypothesis

• Fit a simple linear regression model to each set of (x, y) data, i.e., one model fit to (x₁, y₁), one model fit to (x₂, y₂), one model fit to (x₃, y₃), and one model fit to (x₄, y₄).

• Write down the estimated regression equation for each fitted model, together with the values of the coefficient of determination, r2, and the standard error of the estimate, s=MSE‾‾‾‾‾√.

• For each set of (x, y) data, create a scatterplot of y (vertical) versus x (horizontal) with the estimated regression line added to the plot.

• For each set of (x, y) data, create a scatterplot of the residuals (vertical) versus (horizontal). Based on each plot, do the zero mean and constant variance assumptions about the simple linear regression model error seem reasonable?

• For each set of (x, y) data, create a normal probability plot of the standardized residuals. Based on each plot, does the normality assumption about the simple linear regression model error seem reasonable?

• For each set of (x, y) data, are there any outliers?

• For each set of (x, y) data, are there any high leverage points?

• For each set of (x, y) data, are there any influential points?

  Post a summary of your group’s analysis. What important “big picture” conclusions can you draw from your analysis?

For a population with a mean equal to 250 and a standard deviation equal to 35​, calculate the standard error of the mean for the following sample sizes.

​a) 10

​b)  40

​c)  70

The standard error of the mean for a sample size of 10 is

The standard error of the mean for a sample size of 40 is

The standard error of the mean for a sample size of 70 is

During a blood-donor program conducted during finals week for college students, a blood-pressure reading is taken first, revealing that out of 350 donors, 41 have hypertension. All answers to three places after the decimal. A 95% confidence interval for the true proportion of college students with hypertension during finals week is (WebAssign will check your answer for the correct number of significant figures. , WebAssign will check your answer for the correct number of significant figures. ). We can be 80% confident that the true proportion of college students with hypertension during finals week is WebAssign will check your answer for the correct number of significant figures. with a margin of error of WebAssign will check your answer for the correct number of significant figures. Unless our sample is among the most unusual 10% of samples, the true proportion of college students with hypertension during finals week is between WebAssign will check your answer for the correct number of significant figures. and WebAssign will check your answer for the correct number of significant figures. The probability, at 60% confidence, that a given college donor will have hypertension during finals week is WebAssign will check your answer for the correct number of significant figures. , with a margin of error of WebAssign will check your answer for the correct number of significant figures. Assuming our sample of donors is among the most typical half of such samples, the true proportion of college students with hypertension during finals week is between WebAssign will check your answer for the correct number of significant figures. and WebAssign will check your answer for the correct number of significant figures. We are 99% confident that the true proportion of college students with hypertension during finals week is WebAssign will check your answer for the correct number of significant figures. , with a margin of error of WebAssign will check your answer for the correct number of significant figures. Assuming our sample of donors is among the most typical 99.9% of such samples, the true proportion of college students with hypertension during finals week is between WebAssign will check your answer for the correct number of significant figures. and WebAssign will check your answer for the correct number of significant figures. Covering the worst-case scenario, how many donors must we examine in order to be 95% confident that we have the margin of error as small as 0.01? Using a prior estimate of 15% of college-age students having hypertension, how many donors must we examine in order to be 99% confident that we have the margin of error as small as 0.01?

Thoroughly answer the following questions: What is the importance of generalizability in cost effectiveness analysis?

Suppose that six cards are selected at random from a standard 52-card deck, what is the probability that two of the selected cards are less than 5 and neither of them is black and two cards are greater than 10 neither of them is red, and the last two cards are any cards?

Melanism in Moths. Melanism is the process by which animals produce melanin to darken body tissues and produce color variation. Researchers at Oxford University suspected increased levels of pollution in London may be influencing the evolution of Lepidoptera, a species of moth native to the area that took on one of two distinct camouflage schemes: "light" and "dark." They surveyed a sample of dark and light moths on birch trees and labelled each "conspicuous" if it was visible from 30 yards away or "inconspicuous" if it was not. The table below shows the number of moths of each color that were conspicuous and inconspicuous. Use this data to investigate the researchers' theory that increased levels of pollution might provide an advantage to "dark" moths over "light" ones.

Conduct a test for the stated null and alternative hypotheses. Use ?α = and round numeric answers to four decimal places.

?0H0: The variables use of color and detectability are independent.

??HA: The variables use of color and detectability are not independent.

1. Calculate the test statistic for this hypothesis test.  ? z t X^2 F  =

2. Calculate the degrees of freedom for this hypothesis test.

3. Report the p-value for this hypothesis test out to four decimal places.

p-value =

4. Based on the p-value, we have:
A. very strong evidence
B. some evidence
C. extremely strong evidence
D. little evidence
E. strong evidence
F. Reject ?0H0
that the null model is not a good fit for our observed data.

SPSS assignment – Single-sample t test

A researcher hypothesizes that people who listen to classical music have higher

concentration skills than those in the general population. On a standard

concentration test, the overall population mean is 15.5. The researcher gave the

same test to a random sample of 12 individuals who regularly listen to classical

music. Their scores on the test were as follows:

16, 14, 20, 12, 25, 22, 23, 19, 17, 17, 21, 20

Conduct a single sample t test. Should the null hypothesis be rejected? What should

the researcher conclude? Report the results in APA style.

1) Enter the data into SPSS and Name the variable.

2) Once the data are entered and the variable named, select the Analyze tab, and,

from the drop-down menu, Compare Means followed by One Sample T Test.

A dialog box will pop up. Place the Concentration Score variable into the Test

Variable box by utilizing the arrow in the middle of the window. In the small Test

Value box (under the Test Variable box), enter the value of the population mean.

Click OK

The output for the single-sample t test will appear in an output window.

In the top box labeled “One-Sample Statistics” you will be given the N value, Mean,

Standard Deviation, and Standard Error of the Mean.

In the second box labeled “One-Sample Test” you are given the t value, the degrees

of freedom, the obtained alpha level (significance level), Mean Difference, and the

Lower and Upper limits at 95% confidence interval.

* SPSS reports only two-tailed significance levels, so if you have a one-tailed test,

you must divide the significance level (alpha) in half.

If the alpha (significance level) is less than .05, then the result is significant, and you

reject the null hypothesis and accept the alternative hypothesis.

If the alpha (significance level) is greater than or equal to .05, then you fail to reject

the null hypothesis and do not accept the alternative hypothesis.

Twenty laboratory mice were randomly divided into two groups of 10. Each group was fed according to a prescribed diet. At the end of 3 weeks, the weight gained by each animal was recorded. Do the data in the following table justify the conclusion that the mean weight gained on diet B was greater than the mean weight gained on diet A, at the α = 0.05 level of significance? Assume normality. (Use Diet B - Diet A.)

(a) Find t. (Give your answer correct to two decimal places.)


(ii) Find the p-value. (Give your answer correct to four decimal places.)

(b) State the appropriate conclusion.

Reject the null hypothesis, there is significant evidence that diet B had a greater weight gain. Reject the null hypothesis, there is not significant evidence that diet B had a greater weight gain.     Fail to reject the null hypothesis, there is significant evidence that diet B had a greater weight gain. Fail to reject the null hypothesis, there is not significant evidence that diet B had a greater weight gain.

How would a 2-way ANOVA differ from a 1-way ANOVA ?

Question 1

One of the assumptions of the One-Way ANOVA is that the samples for all three (or more) groups are dependent.

True

False

Question 2

One of the assumptions of the One-Way ANOVA is that the samples are from a normal distribution.

True

False

Question 3

Sum of Squares Between + Sum of Squares Within = Sum of Squares Total

Breaking the Sum of Squares Total up into SS_Between and SS_Within is called ____________________ the sum of squares.

answer choices:

seperating

fractions

partitioning

dividing

Question 4

The test statistic for ANOVA's come from the ____ distribution.

answer choices:

F

n

t

z

Question 5

The numerator for the F-statistic is _____________________.

answer choices:

Mean Square Within

Sum of Squares Between

Sum of Square Within

Mean Square Between

Question 6

The formula for degrees of freedom total is _________________________.

answer choices:

n-k

k-1

N-1

n-1

Question 7

The degrees of freedom between is ______________________.

answer choices:

N-1

k-1

n-k

n-1

Question 8

Sum of Squares ________________ is calculated by subtracting each group mean from the grand total mean, squaring each of these differences, and adding these squared-differences together.

answer choices:

Total

Means

Within

Between

Question 9

Sum of Squares __________________ is calculated by subtracting every single score from the grand total mean, squaring each of t

hese differences, and adding all of these squared-differences together.

answer choices:

Total

Between

Scores

Within

Question 10

The alternative hypothesis for a One-Way Anova predicts M₁ = M₂ = M₃ = . . . = M_k, with k being the number of different groups in our data set.

answer choices:

True

False