Question 1

Flu cases per school this past flu season in your parish school system (\(n=500\)) were 15 per week. For the entire state, the weekly mean was 16, and the standard deviation was 15.1. Are the kids in your parish as sick as the kids throughout the state? Set the significance level to 0.05.

Proof:

Because we have a large sample size, \(n=500\), and we also have the state data that represents the population statistics, we can use a Z-Test to answer our question. For this, we’ll utilize the 6-Step Hypothesis process.

  1. Identify the Population & Sample

    1. Population is the number of flu cases in the State.

    2. Sample is the number of flu cases in the parish school system.

  2. State the Null and Alternative Hypothesis

    1. \(H_0: \mu = 16\)

    2. \(H_\alpha: \mu \neq 16\)

      • This implies that we are dealing with a two-tailed, non-directional test.

  3. State the Assumptions and Check the Conditions

    1. We have the following information to perform the test:

      • \(\mu = 16\)

      • \(\bar{x}=15\)

      • \(n=500\)

      • \(\sigma = 15.1\)

  4. Calculate the Relevant Test-Statistic (Z-Test Value)

\[ z = \frac{\bar{x}-\mu_0}{\sigma_{\bar{x}}}= \frac{\bar{x}-\mu_0}{\frac{\sigma}{\sqrt{n}}} \]

PopMean1 <- 16
SampMean1 <- 15
SampSize1 <- 500
Std1 <- 15.1
z1 <- (SampMean1 - PopMean1)/(Std1/sqrt(SampSize1))
print(z1)
## [1] -1.48084
  1. Calculate the P-Value & Critical Values
p1<- 2*pnorm(-abs(z1))
print(p1)
## [1] 0.1386493
critical1 <- qt(p=.05/2, df=499, lower.tail=FALSE)
critical1a <- -qt(p=.05/2, df=499, lower.tail=FALSE)
print(critical1)
## [1] 1.964729
print(critical1a)
## [1] -1.964729
  1. State the Conclusion
    • Because the \(p\)-value is greater than our significance level, \(0.05\), we will fail to reject the null hypothesis.
    • In addition, the \(z\)-statistic is greater than our critical value, means that it falls in our safety zone, indicating that we will fail to reject our null hypothesis.

QED

Question 2

Use a 0.05 significance level to test the claim that college students have a mean IQ of 110. The sample data are summarized by the statistics \(n=25\), \(\bar{x}=118.0\), and \(s=10.7\)

Proof:

Because we have the sample standard deviation, we will be performing a T-Test because we have the standard deviation of the sample, not the population. For this, we’ll utilize the 6-Step Hypothesis process.

  1. Identify the Population & Sample

    1. Population is the number of college students.

    2. Sample is the 25 students that were surveyed.

  2. State the Null and Alternative Hypothesis

    1. \(H_0: \mu = 110\)

    2. \(H_\alpha: \mu \neq 110\)

      • This implies that we are dealing with a two-tailed, non-directional test.

  3. State the Assumptions and Check the Conditions

    1. We have the following information to perform the test:

      • \(\mu = 110\)

      • \(\bar{x}=118\)

      • \(n=25\)

      • \(s=10.7\)

  4. Calculate the Relevant Test-Statistic (T-Test Value)

    \[ t = \frac{\bar{x}-\mu_0}{s_{\bar{x}}}= \frac{\bar{x}-\mu_0}{\frac{s}{\sqrt{n}}} \]

muknot <- 110
xbar <- 118
SampSize2 <- 25
Std2 <- 10.7
t2 <- (xbar - muknot)/(Std2/sqrt(SampSize2))
print(t2)
## [1] 3.738318
  1. Calculate the P-Value & Critical Value
p2<- 2*pnorm(-abs(t2))
print(p2)
## [1] 0.0001852557
critical2 <- qt(p=.05/2, df=24, lower.tail=FALSE)
print(critical2)
## [1] 2.063899
critical2a <- -qt(p=.05/2, df=24, lower.tail=FALSE)
print(critical2a)
## [1] -2.063899
  1. State the Conclusion
    • Because the test statistic is greater than our critical value, meaning that it falls in the rejection zone, we will reject the null hypothesis.
    • In addition, because the \(p\)-value is less than our \(\alpha\) value, we will reject the null hypothesis.

QED

Question 3

Use a 0.05 significance level to test the claim that the mean life span of cell phones is equal to 5 years. The sample data is summarized by the statistics: \(n=27\), \(\bar{x}=4.6\) years, and \(s=1.9\) years.

Proof:

Because we have the sample standard deviation, we will be performing a T-Test because we have the standard deviation of the sample, not the population. For this, we’ll utilize the 6-Step Hypothesis process.

  1. Identify the Population & Sample

    1. Population is the mean life span of all cell phones

    2. Sample is the mean life span of the 27 cell phones that were sampled.

  2. State the Null and Alternative Hypothesis

    1. \(H_0: \mu = 5\)

    2. \(H_\alpha: \mu \neq 5\)

      • This implies that we are dealing with a two-tailed, non-directional test.

  3. State the Assumptions and Check the Conditions

    1. We have the following information to perform the test:

      • \(\mu_0 = 5\)

      • \(\bar{x}=4.6\)

      • \(n=27\)

      • \(s=1.9\)

  4. Calculate the Relevant Test-Statistic (T-Test Value)

    \[ t = \frac{\bar{x}-\mu_0}{s_{\bar{x}}}= \frac{\bar{x}-\mu_0}{\frac{s}{\sqrt{n}}} \]

muknot3 <- 5
xbar3 <- 4.6
SampSize3 <- 27
Std3 <- 1.9
t3 <- (xbar3 - muknot3)/(Std3/sqrt(SampSize3))
print(t3)
## [1] -1.093927
  1. Calculate the P-Value & Critical Value
p3<- 2*pnorm(-abs(t3))
print(p3)
## [1] 0.2739871
critical3 <- qt(p=.05/2, df=26, lower.tail=FALSE)
print(critical3)
## [1] 2.055529
critical3a <- -qt(p=.05/2, df=26, lower.tail=FALSE)
print(critical3a)
## [1] -2.055529
  1. State the Conclusion
    • Because the T-statistic falls between -2.05 and +2.05, that means that it also falls in the safety zone, indicating that we will fail to reject the null hypothesis.
    • In addition, the \(p\) value is greater than our \(\alpha\) value, indicating that we will fail to reject the null hypothesis.

QED

Question 4

The population of Children Behavior Problems scores has a mean score of 50 and a variance of 100. Assume that we have a sample of fifteen children and have a mean scores on Children Behavior Problems of 56. Test the claim that the mean of these fifteen children is equal to the mean from a population of normal children at the 0.05 level of the significance level.

Proof:

Because we have the data that represents the population (specifically population variation, implying population standard deviation), we can use a Z-Test to answer our question. For this, we’ll utilize the 6-Step Hypothesis process.

  1. Identify the Population & Sample

    1. Population is the scores of Children Behavior Problems

    2. Sample is the 15 Children chosen from the population

  2. State the Null and Alternative Hypothesis

    To correctly label, let’s say that Mean of the Sample of 15 Children is \(\mu_{sc}\) and the Mean of the Population of Children Behavior Problem scores is \(\mu_{pc}\)

    1. \(H_0: \mu_{sc}=\mu_{pc}\)

    2. \(H_\alpha: \mu_{sc}\neq \mu_{pc}\)

      • This implies that we are dealing with a two-tailed, non-directional test.

  3. State the Assumptions and Check the Conditions

    1. We have the following information to perform the test:

      • \(\mu = 50\)

      • \(\bar{x}=56\)

      • \(n=15\)

      • \(\sigma = 10\)

  4. Calculate the Relevant Test-Statistic (Z-Test Value)

\[ z = \frac{\bar{x}-\mu_0}{\sigma_{\bar{x}}}= \frac{\bar{x}-\mu_0}{\frac{\sigma}{\sqrt{n}}} \]

PopMean4 <- 50
SampMean4 <- 56
SampSize4 <- 15
Std4 <- 10
z4 <- (SampMean4 - PopMean4)/(Std4/sqrt(SampSize4))
print(z4)
## [1] 2.32379
  1. Calculate the P-Value & Critical Value
p4<- 2*pnorm(-abs(z4))
print(p4)
## [1] 0.02013675
critical4 <- qt(p=.05/2, df=14, lower.tail=FALSE)
print(critical4)
## [1] 2.144787
critical4a <- -qt(p=.05/2, df=14, lower.tail=FALSE)
print(critical4a)
## [1] -2.144787
  1. State the Conclusion
    • Because the \(p\)-value is less than our significance level, \(0.05\), we will reject the null hypothesis.
    • In addition, the test statistic is greater than our critical value, indicating that falls in the rejection zone, implying that we will reject the null hypothesis.

QED

Notes and Reminders

P-Values

  1. If \(p > \alpha\) , then we will Fail to Reject the Null Hypothesis
  2. If \(p<\alpha\), then we will Reject the Null Hypothesis

Critical Values

  1. If the Test Statistic falls outside of the critical value interval, then we will Reject the Null Hypothesis

  2. If the Test Statistic is between the two critical values, then we will Fail to Reject the Null Hypothesis

Z-Test vs T-Test

  1. Z-Tests only occur when we have a Population Standard Deviation or Population Variance

  2. T-Tests only occur when only have a Sample Standard Deviation or Sample Variance. No Population STD or Variance is given.