Correlation Analysis

Question 1

A cookie company is evaluating a new machine that should produce cookies with a mean breaking strength of 70 pounds and a standard deviation of 3.5 pounds. A sample of 49 cookies reveals a sample mean breaking strength of 69.1 pounds. The questions we need to address are:

A. State the null and alternative hypotheses.
B. Is there evidence the machine is not meeting the manufacturer’s specifications at a 0.05 significance level?
C. Compute the p-value and interpret its meaning.
D. What if the standard deviation were 1.75 pounds?
E. What if the sample mean were 69 pounds?

A. State the null and alternative hypothesis _______

The null and alternative hypotheses are:

Null Hypothesis (H₀): The mean breaking strength of cookies is 70 pounds. H₀: μ = 70

Alternative Hypothesis (H₁): The mean breaking strength of cookies is not 70 pounds. H₁: μ != 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 _______

We perform a two-tailed hypothesis test at a 0.05 significance level (α = 0.05). First, we calculate the z-score:

z = (x̄ - μ) / (σ / √n)

Where: x̄ = 69.1 (sample mean) μ = 70 (population mean) σ = 3.5 (standard deviation) n = 49 (sample size)

Substitute the values: z = (69.1 - 70) / (3.5 / √49) z = -0.9 / 0.5 z = -1.8

The critical z-values for a 0.05 significance level are +- 1.96. Since the absolute value of z (|z| = 1.8) is less than 1.96, we do not reject the null hypothesis. This suggests there is insufficient evidence to conclude that the machine is not meeting the manufacturer’s specifications.

C. Compute the p value and interpret its meaning _______

The p-value is calculated using the R function pnorm:

p_value <- 2 * pnorm(-abs(-1.8))
p_value

The result is approximately 0.0718. Since the p-value is greater than 0.05, we do not reject the null hypothesis.

At the 0.05 significance level, there is no evidence to suggest that the machine is producing cookies with a mean strength different from 70 pounds.

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

We recalculate the z-score with the new standard deviation:

z = (x̄ - μ) / (σ / √n)

Substitute the values: z = (69.1 - 70) / (1.75 / √49) z = -0.9 / 0.25 z = -3.6

The p-value is:

p_value <- 2 * pnorm(-abs(-3.6))
p_value

The p-value is much smaller than 0.05, so we reject the null hypothesis.

With a standard deviation of 1.75 pounds, there is strong evidence that the machine is not meeting the manufacturer’s specifications.

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

We adjust the sample mean to 69 pounds and recalculate the z-score:

z = (x̄ - μ) / (σ / √n)

Substitute the values: z = (69 - 70) / (3.5 / √49) z = -1 / 0.5 z = -2

The p-value is:

p_value <- 2 * pnorm(-abs(-2))
p_value

The p-value is approximately 0.0455. Since this is less than 0.05, we reject the null hypothesis.

With a sample mean of 69 pounds, we conclude that the machine is not meeting the manufacturer’s specifications.

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

Let’s calculate the 95% confidence interval for the population mean in a separate case where:

Sample mean (x̄) = 85
Standard deviation (σ) = 8
Sample size (n) = 64

For a 95% confidence level, the critical z-value is 1.96. Using the formula:

CI = x̄ ± z* × (σ / √n)

Substitute the values: CI = 85 ± 1.96 × (8 / √64) CI = 85 ± 1.96 × 1 CI = 85 ± 1.96

The confidence interval is: (83.04, 86.96)

We are 95% confident that the true population mean lies between 83.04 and 86.96.

Question 3, using Correlation Analysis

girls_data <- data.frame(
  Goals = c(4, 5, 6),
  Grades = c(49, 50, 69),
  Popular = c(24, 36, 38),
  Time_Spent = c(19, 22, 28),
  Total = c(92, 108, 135)
)

boys_data <- data.frame(
  Goals = c(4, 5, 6),
  Grades = c(46.1, 54.2, 67.7),
  Popular = c(26.9, 31.6, 39.5),
  Time_Spent = c(18.9, 22.2, 27.8),
  Total = c(95.9, 113, 141)
)


print("girl data:")

print(girls_data)

print("boy data:")

print(boys_data)

Result:[1] "girl data:"

  Goals Grades Popular Time_Spent Total
1     4     49      24         19    92
2     5     50      36         22   108
3     6     69      38         28   135
[1] "boy data:"
  Goals Grades Popular Time_Spent Total
1     4   46.1    26.9       18.9  95.9
2     5   54.2    31.6       22.2 113.0
3     6   67.7    39.5       27.8 141.0