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I NEED THE SOLUTIONS TO F, G, H, I, and J \"lensthickness.csv\" file shown below

ID: 3043334 • Letter: I

Question

I NEED THE SOLUTIONS TO F, G, H, I, and J

"lensthickness.csv" file shown below:

obs,thickness
1,2.92
2,2.73
3,3.31
4,3.28
5,3.2
6,2.82
7,3.27
8,2.39
9,3.41
10,3.31
11,3.32
12,3.22
13,3.35
14,3.41
15,2.69
16,3.29
17,3.33
18,3.25
19,3.15
20,2.84
21,2.8
22,3.15
23,2.79
24,3.49
25,3.17
26,2.85
27,3.05
28,3.08
29,3.34
30,3.01
31,3
32,3.31

Part V. (17 points) Hypothesis Testing in Practice. A manufacturer of eyeglass lenses routinely checks the thickness of their lenses to ensure they are approximately 3.20 mm, which is the desired lens thickness. A sample of 32 lenses, chosen at random, from a larger batch yields the data provided in lesthickness.csv. If the observed data provides evidence the average thickness of the batch is other than 3.20 mm, the manufacturing process will need to be investigated and possibly reconfigured Is there enough evidence to assume the average lens thickness, from the batch, is different than the desired 3.20mm? Use the dataset, lensthickness.csv located in the Data Analysis #3 tab in canvas, to answer the following questions You will need to write your own R code to complete the analysis a. (1 point) What is the parameter of interest in this scenario? Provide the symbol and context. b. (1 point) State the null and alternative hypothesis to answer the question of interest. c. (1 point) Create an appropriate plot in R to visualize the distribution of lens thickness include a title. (You may refer to lessons from week 3 or DA#2 for code to create a histogram or boxplot of the data or any other R reference materials.) Paste your plot. d. (1 point) Is there visual evidence that the true mean is other than 3.20 mm? Explain. Must include context. e. (1 point) Calculate the sample mean and standard deviation using R. State the values. (You may refer to lessons f. g. h. i. j. from week 3 or DA#2 for code to create a histogram or boxplot ofthe data or any other R reference materials) (2 point) Check the conditions for inference. State them and whether they are met. (2 points) Calculate the test statistic. Show work. (1 point) State the p-value. Is it one or two sided? (2 points) Calculate the 95% Confidence Interval. Show work. (1 point)Use the ttest() command in R to verify the results of the t test. See code in Data Analysis #3 tab in camvas. How do your answers compare? k. (4 points) From the R output, write a four-part conclusion describing the results. Use = 0.05. Provide a statement in terms ofthe alternative hypothesis. State whether or not to reject the nul1. Give your interpretation of the point and interval estimate, in context, and to the best of your ability. Include any other information you might feel to relevant. Append your code 1.

Explanation / Answer

g.Test statistic : T = n * [mean(Thickness) – 3.2] / sd(thickness) , where n = sample size
                               = 32 * (3.110313-3.2) / 0.260638 = -1.946555

h. p-value = P H0 [ |T| > |-1.946555| ]
                 = P [ |t n-1| > 1.946555 ], since T ~ t n-1 under H0 : mean = 3.2, n=32
                   = 2. P [ t 31 > 1.946555 ]
                  = 2 * 0.0303497 = 0.06069939
This is two sided, since here we have alternative hypothesis H1 : mean 3.2

i. Confidence interval : mean(thickness) ± t 0.025;31 * sd(thickness) / n
= ( 3.110313 ± t 0.025;31 * 0.260638 / 32 )
= ( 3.01634199 , 3.204283211 )

j. > t.test(thickness,mu=3.2)

        One Sample t-test

data: thickness

t = -1.9466, df = 31, p-value = 0.0607

alternative hypothesis: true mean is not equal to 3.2

95 percent confidence interval:

3.016342 3.204283

sample estimates:

mean of x

3.110313