Question

A researcher compares two compounds (1 and 2) used in the manufacture of car tires that are designed to reduce braking distances for SUVs equipped with the tires. SUVs equipped with tires using compound 1 have a mean braking distance of 62 feet and a standard deviation of 10.5 feet. SUVs equipped with tires using compound 2 have a mean braking distance of 66 feet and a standard deviation of 12.0 feet. Suppose that a sample of 86 braking tests are performed for each compound. Using these results, test the claim that the braking distance for SUVs equipped with tires using compound 1 is shorter than the braking distance when compound 2 is used. Let μ1 be the true mean braking distance corresponding to compound 1 and μ2 be the true mean braking distance corresponding to compound 2. Use the 0.05 level of significance.

Answer 1

Answer:

$t=\frac{62-66}{\sqrt{\frac{10.5^2}{86}+\frac{12^2}{86}}}}=-2.326$  

$p_v =P(t_{170}$

Comparing the p value with the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we can reject the null hypothesis, so the mean for the compound 1 is significantly less than the mean for the compound 2.

Step-by-step explanation:

Data given and notation

$\bar X_{1}=62$ represent the mean for the compound 1

$\bar X_{2}=66$ represent the mean for the compound 2

$s_{1}=10.5$ represent the sample standard deviation for compound 1

$s_{2}=12$ represent the sample standard deviation for compund 2

$n_{1}=86$ sample size for the group compound 1

$n_{2}=86$ sample size for the group compound 2

t would represent the statistic (variable of interest)

$\alpha=0.05$ significance level provided

Develop the null and alternative hypotheses for this study?

We need to conduct a hypothesis in order to check if the means for compound 1 is less than the mean for compound 2, the system of hypothesis would be:

Null hypothesis:$\mu_{1}\geq\mu_{2}$

Alternative hypothesis:$\mu_{1} < \mu_{2}$

Since we don't know the population deviations for each group, for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{\bar X_{1}-\bar X_{2}}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}}$ (1)

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

Determine the critical value(s).

Based on the significance level$\alpha=0.05$, we can find the critical values with the t distribution with 86+86-2=170 degrees of freedom, we are looking for values that accumulates 0.05 of the left tail of the distirbution.

For this case the value is $t_{\alpha}=-1.653$

Calculate the value of the test statistic for this hypothesis testing.

Since we have all the values we can replace in formula (1) like this:

$t=\frac{62-66}{\sqrt{\frac{10.5^2}{86}+\frac{12^2}{86}}}}=-2.326$  

What is the p-value for this hypothesis test?

Since is a left tailed test the p value would be:

$p_v =P(t_{170}$

Based on the p-value, what is your conclusion?

Comparing the p value with the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we can reject the null hypothesis, so the mean for the compound 1 is significantly less than the mean for the compound 2.