lucianolopezc06-sbnlu

this was my module 5 assignment

Question 1

A. Null hypothesis: H0: u = 70

 Alternative hypothesis: H1: u is not equal to 70

 B. z = (69.1 – 70) / (3.5 / sqrt(49)) z = -1.80

 Since -1.80 is between -1.96 and 1.96, we fail to reject the null hypothesis. There is not enough evidence that the machine does not meet the specifications.

 C. p-value = 0.0719

 Since 0.0719 > 0.05, we fail to reject the null hypothesis. There is not enough evidence that the average breaking strength is different from 70 pounds.

D. z = (69.1 – 70) / (1.75 / sqrt (49))

z = -3.60

We reject the null hypothesis. There is enough evidence that the machine does not meet the specifications.

 E. z = (69 – 70) / (3.5 / sqrt (49))

z = -2.00

Since -2.00 < -1.96, we reject the null hypothesis. There is enough evidence that the machine does not meet the specifications.

Question 2

x bar = 85

 sigma = 8

n = 64

SE = 8 / sqrt(64) = 1

CI = 85 +/- 1.96(1)

CI = 85 +/- 1.96

 95% CI = (83.04, 86.96)

Question 3

# Girls data

x1 <- c(4, 5, 6)

x2 <- c(19, 22, 28)

# Boys data

y1 <- c(4, 5, 6)

y2 <- c(18.9, 22.2, 27.8)

# Create data frame

df <- data.frame(x1, x2, y1, y2)

# Correlation

cor(df)

# Pearson correlation

cor(df, method = “pearson”)

# Spearman correlation

cor(df, method = “spearman”)

# Plot of the correlation

plot(x1, x2, main = “Correlation Between Goals and Time Spent”,

     xlab = “Goals”,

     ylab = “Time Spent on Assignment”)

# Correlogram

library(corrgram)

corrgram(df)

The Pearson correlation between girls’ goals and time spent on the assignment is 0.982, showing a strong positive relationship and the plot visually represents this.