Chapter 11, Problem 3WAI

Describe the differences between performing a hypothesis test for ${\mu }_1-{\mu }_2$ with paired samples and performing the same test for independent samples.

To determine

To explain:

The differences between performing a hypothesis test for ${\mu }_1-{\mu }_2$ with paired samples and performing the same test for independent samples.

Answer to Problem 3WAI

The p-value for independent-samples $t$ test is $0.066$ higher than the paired-samples $t$ test $0.012$ due to independent-samples $t$ -test is each sample is taken from different populations.

Explanation of Solution

Given Information:

To test the hypothesis is that the mean emissions are higher at $40$ degree at $5\%$ significance level.

The null and alternative hypothesis is,

  $\begin{array}{l}{H}_0:{\mu }_d=0\\ H_1:{\mu }_d>0\end{array}$

Calculation:

Paired $T$ -Test and $\text{CI}:{40}^{\text{o}}F,{80}^{\text{o}}F$ .

Paired $T$ for ${40}^{\text{o}}F-{80}^{\text{o}}F$

  $\begin{array}{l}\text{ N Mean StDev SE Mean}\\ {\text{40}}^{\text{o}}F10\text{ 821}\text{.3 59}\text{.9 18}\text{.8}\\ {\text{80}}^{\text{o}}F\text{ 10 785}\text{.0 40}\text{.6 12}\text{.8}\\ \text{Difference 10 36}\text{.3 42}\text{.3 }\text{13}\text{.4}\end{array}$

  $\text{95\%}$ Lower bound for mean difference: $11.8$

  $\text{T}$ Test of mean Difference $=0\left(\text{vs }>0\right):\text{T-Value}=2.71\text{ p-Value}=0.012$

  $2.71$

From the MINITAB output, the t-test statistics is,

The p-value for this test is,

  $0.012$

The conclusion is that $p$ -Value in this context is less than $0.05$ which is $0.012$ , so the null hypothesis is rejected at $5\%$ level of significance. There is sufficient evidence to indicate that the mean emissions are higher at $40$ degree. The result is statistically significant.

To test the hypothesis is that the mean emissions are higher at $40$ degree at $5\%$ significance level.

The null and alternative hypothesis is,

  $\begin{array}{l}{H}_0:{\mu }_1={\mu }_2\\ H_a:{\mu }_1>{\mu }_2\end{array}$

The t-test statistics is,

By using MINITAB, friend $t$ -test statistics with the help of the following steps

1. Import the data.
2. Select me Stat and choose the Basic Statistics option.
3. Select me $2$ sample $t$ and enter Summarized data
4. Click option button choose Confidence level, Test difference and alternative hypothesis
5. Click 0k.

Two Sample T-Test and CI: ${40}^{\text{o}}F,{80}^{\text{o}}F$

Two sample $T\text{ for }{40}^{\text{o}}F\text{ vs }{80}^{\text{o}}F$ .

  $\begin{array}{l}\text{ N Mean StDev SE Mean}\\ {\text{40}}^{\text{o}}F10\text{ 821}\text{.3 59}\text{.9 }\text{19}\\ {\text{80}}^{\text{o}}F\text{ 10 785}\text{.0 40}\text{.6 13}\end{array}$

Difference $=mu\left({40}^{\text{o}}F\right)=mu\left({80}^{\text{o}}F\right)$

Estimate for difference: $36.3$

  $95\%$ Lower bound for difference: $-3.6$

  $T$ -Test of difference $=0\left(\text{vs >}\right)$ : $\text{T}$ Value $=1.59,\text{ p-Value}=0.066\text{ DF}=15$

  $1.59$ From the MINITAB output, the $t$ test statistics is,

The p-value for this test is,

  $0.066$

The conclusion is that $p$ -value in this context is higher than $\text{0}\text{.05 which is 0}\text{.066}$ , so the null hypothesis is not rejected at $\text{5\%}$ level of significance. There is insufficient evidence to indicate that the mean emissions are higher at $40$ degree. The result is not statistically significant.

The $p$ -value for independent-samples $t$ test is $0.066$ higher than the paired-samples $t$ test $0.012$ due to independent-samples $t$ -test is each sample is taken from different populations.