Module # 5 Assignment

 Question 1:

The director of manufacturing at a cookies company needs to determine whether a new machine is able to produce a particular type of cookies according to the manufacturer's specifications, which indicate that cookies should have a mean of 70 and standard deviation of 3.5 pounds. A sample of 49 cookies reveals a sample mean breaking strength of 69.1 pounds.



This screenshot covers part A and part B




A. State the null and alternative hypothesis 

The null hypothesis is that the new machine will be able to produce cookies with a mean of 70. The alternative hypothesis is that the new machine will not be able to produce cookies with the mean breaking strength of 70. 


B. Is there evidence that the machine is not meeting the manufacturer's specifications for
average strength? Use a 0.05 level of significance 

SE = (3.5 / √49) = 0.5
z = (69.1- 70) / (0.5) = -0.9 / 0.5 = -1.8




This screenshot covers part C
C. Compute the p value and interpret its meaning 

The p value is 0.07186064. This value is greater than the alpha value so the null hypothesis is not rejected. This is due to not being enough evidence that the machine is not meeting the manufacturer’s specifications.








This screenshot covers part D

D. What would be your answer in (B) if the standard deviation were specified as 1.75 pounds?


z = 69.1 -70 / (1.75/√49) = -0.9 / (1.75/7) = -0.9 /0.25 = -3.6

The new p-value is 0.0003182172. Due to the new p-value being less than the level of significance of 0.05, the null hypothesis is rejected.






This screenshot covers part E


E. What would be your answer in (B) if the sample mean were 69 pounds and the standard deviation is 3.5 pounds? 


z = 69 - 70 / (3.5/√49) = -1 / 0.5 = -2

The new p-value is 0.04550026. Due to the new p-value being less than the level of significance of 0.05, the null hypothesis is rejected.







Question 2:
If x̅ = 85, σ = standard deviation = 8, and n=64, set up 95% confidence interval estimate of the population mean μ.
  








The Lower bound is 83.04 and the Upper bound is 86.96





Question 3, using Correlation Analysis:
Using the dataset downloadable below (i.e. the data you will use to create the vectors are located in the download link below), complete these tasks in Rstudio:


First part of code in screenshots below


a. Calculate the correlation coefficient for this data set 

b. Pearson correlation coefficient






cor(df)
cor(df, method = "pearson")
cor(df, method = "spearman")
c. Create plot of the correlation











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