Seven less than the sum of p and t is as much as 6

.... pertains to the first choice

Answer:

There are 369 students who enrolled in either calculus or discrete mathematics.

Step-by-step explanation:

I'll create a Venn diagram to illustrate these figures.

Let’s define:

A represents the count of students who completed calculus.

B stands for those who took discrete mathematics.

We have the following:

Here, a denotes the students who studied calculus exclusively, while represents those who took both subjects.

Using the same reasoning:

There are 188 students taking both calculus and discrete mathematics.

This implies that

A total of 212 students studied discrete mathematics.

<pBased on this,

345 students have attended a calculus course.

<pSo, from this information,

We find the total number of students who took either calculus or discrete mathematics.

A total of 369 students participated in either calculus or discrete mathematics.

Response:

Null hypothesis = H₀ = σ₁² ≤ σ₂²

Alternative hypothesis = Ha = σ₁² > σ₂²

Calculated statistic = 1.9

p-value = 0.206

Given that the p-value exceeds α, we do not reject the null hypothesis.

Thus, we conclude that night shift workers do not exhibit higher variability in their output levels compared to day workers.

Step-by-step elaboration:

Let σ₁² represent the variance for night shift workers

Let σ₂² represent the variance for day shift workers

State the null and alternative hypotheses:

The null hypothesis suggests that the variance of night shift workers does not exceed that of day shift workers.

Null hypothesis = H₀ = σ₁² ≤ σ₂²

The alternative hypothesis posits that the variance for night shift workers surpasses that of day shift workers.

Alternative hypothesis = Ha = σ₁² > σ₂²

Calculated statistic:

The test statistic, or F-value, is derived using

Test statistic = Larger sample variance/Smaller sample variance

The larger sample variance is σ₁² = 38

The smaller sample variance is σ₂² = 20

Test statistic = σ₁²/σ₂²

Test statistic = 38/20

Calculated statistic = 1.9

p-value:

The corresponding degrees of freedom for night shift workers is[1]

df₁ = n - 1

df₁ = 9 - 1

df₁ = 8

The corresponding degrees of freedom for day shift workers is[1]

df₂ = n - 1

df₂ = 8 - 1

df₂ = 7

We can obtain the p-value using the F-table or Excel.

To determine the p-value in Excel, we use

p-value = FDIST(F-value, df₁, df₂)

p-value = FDIST(1.9, 8, 7)

p-value = 0.206

Conclusion:

p-value > α    

0.206 > 0.05   ( α = 0.05)As the

p-value is larger than α, we do not reject the null hypothesis with a confidence level of 95%

[[TAG_101]]This leads us to conclude that night shift workers do not demonstrate more variability in their output levels in comparison to day workers.[[TAG_102]]