Module #5 Assignment

 9/25/2023


1. The director of manufacturing at a cookies needs to determine whether a new machine is production 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 pf 49 of cookies reveals a sample mean breaking strength of 69.1 pounds.
A. State the null and alternative hypothesis 

H0: The machine is producing cookies in line with the manufacturer's specifications.

H1: The machine is not producing cookies in line with the manufacturer's specifications.


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

For this I used the z test and ended up with the value of -1.4. After determining critical values, -1.4 is within the range. Therefore, we fail to reject the null hypothesis. 

C. Compute the p value and interpret its meaning

p value = 0.1628

The p value is greater than the significance value, therefore we fail to reject the null hypothesis.


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

In this case, the z statistic comes out to -3.77. Since this is far outside the critical value range, we reject the null hypothesis. So in this case, the machine may not be meeting the manufacturer's specifications for strength.

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

These values match the manufacturer's specifications and therefore will likely not reject the null hypothesis. 

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

a. Z= 1.96

b. plug in limits

confidence interval = 83.05, 86.96

3. a. Calculate the correlation coefficient for this data set 

    In R, I got this answer: [1] 0.9441243

   b. Pearson correlation coefficient 

    [1] 0.8873565

   c. Create plot of the correlation


Comments

Post a Comment

Popular posts from this blog

Module #11 Assignment

 11/6/2023 10.1 library(ISwR) data(ashina) data ashina$subject <- factor(1:16) attach(ashina) act <- data.frame(vas=vas.active, subject, treat=1, period=grp) plac <-data.frame(vas=vas.plac, subject, treat=0, period = grp) first I created the model:  additive_model <- aov(vas ~ treat + subject + period, data = rbind(act, plac)) > anova(additive_model) Analysis of Variance Table Response: vas Df Sum Sq Mean Sq F value Pr(>F) treat 1 14706 14706.1 10.413 0.005644 ** subject 15 51137 3409.2 2.414 0.049184 * Residuals 15 21184 1412.3 --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 > t_test_1 <- t.test(vas ~ treat, data = ashina, subset = period == 1) > t_test_2 <- t.test(vas ~ treat, data = act, subset = period == 2) 10.3 > model_1 <- lm(z ~ a * b) model_2 <- lm(z ~ a:b) Call: lm(formula = z ~ a * b) Coefficients: (Intercept) ...
Read more

Module #12 Assignment

Image
First, I assigned the credit card data to a data frame.: credit_card_data <- data.frame(   Month = c("Jan", "Feb", "Mar", "Apr", "May", "Jun", "Jul", "Aug", "Sep", "Oct", "Nov", "Dec"),   `2012` = c(31.9, 27, 31.3, 31, 39.4, 40.7, 42.3, 49.5, 45, 50, 50.9, 58.5),   `2013` = c(39.4, 36.2, 40.5, 44.6, 46.8, 44.7, 52.2, 54, 48.8, 55.8, 58.7, 63.4)    Next I called on ggplot 2 and created the time series plot: library(ggplot2) credit_card_ts <- ts(t(credit_card_data[, -1]), start = c(2012, 1), frequency = 12) matplot(credit_card_ts, type = "l", lty = 1, col = 1:ncol(credit_card_ts),         xlab = "Time", ylab = "Charges", main = "Monthly Credit Card Charges") Next, I employed the exponential soothing model: credit_card_model <- HoltWinters(credit_card_ts, beta = FALSE, gamma = FALSE) print(credit_card_model) Holt-Winters e...
Read more

Module #9 Assignment

  1. Your data.frame is > assignment_data <- data.frame( Country = c("France","Spain","Germany","Spain","Germany", "France","Spain","France","Germany","France"), age = c(44,27,30,38,40,35,52,48,45,37), salary = c(6000,5000,7000,4000,8000), Purchased=c("No","Yes","No","No","Yes", "Yes","No","Yes","No","Yes")) Generate simple table in R that consists of four rows: Country, age, salary and purchased. In order to make a table from the dataframe, I loaded the data.table package. This is the result: > setDT(assignment_data) > assignment_data Country age salary Purchased 1: France 44 6000 No 2: Spain 27 5000 Yes 3: Germany 30 7000 No 4: Spain 38 4000 No 5: Germany 40 8000 Yes 6: France 35 6000 Yes 7: Spain 52...
Read more