Question

Engineers at a large automobile manufacturing company are trying to decide whether to purchase brand A or brand B tires for the company’s new models. To help them arrive at a decision, an experiment is conducted using 12 of each brand. The tires are run until they wear out. The results are as follows: Brand A: Brand B:x¯¯¯1=37,900 kilometers, s1=5100 kilometers. x¯¯¯1=39,800 kilometers, s2=5900 kilometers. Test the hypothesis that there is no difference in the average wear of the two brands of tires. Assume the populations to be approximately normally distributed with equal variances. Use a P-value.

Answer 1

Answer:

$s_p =\sqrt{\frac{(12 -1)(5100)^2 +(12-1)(5900)^2}{12 +12 -2}}=5514.526$

$t=\frac{(37900-39800)-0}{5514.526\sqrt{\frac{1}{12}+\frac{1}{12}}}}=-0.844$  

$p_v =2*P(t_{22}$  

Comparing the p value with a significance level for example $\alpha=0.1$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can conclude that the true means are not significantly different.  

Step-by-step explanation:

Data given and notation

$\bar X_{A}=37900$ represent the mean for A

$\bar X_{B}=39800$ represent the mean for B

$s_{A}=5110$ represent the sample standard deviation for A

$s_{B}=5900$ represent the sample standard deviation for B

$n_{A}=12$ sample size for the group A  

$n_{B}=12$ sample size for the group B

$\alpha$ Significance level provided  

t would represent the statistic (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to check if the means are equal or not, the system of  hypothesis would be:  

Null hypothesis:$\mu_{A}-\mu_{B}= 0$  

Alternative hypothesis:$\mu_{A} - \mu_{B}\neq 0$  

We don't have the population standard deviation's but we assume that the population deviation is equal for both populations, so we can apply a t test to compare means, and the statistic is given by:  

$t=\frac{(\bar X_{A}-\bar X_{B})-\Delta}{s_p\sqrt{\frac{1}{n_{A}}+\frac{1}{n_{B}}}}$ (1)  

Where $s_p$ represent the standard deviation pooled given by:

$s_p =\sqrt{\frac{(n_A -1)s^2_A +(n_B -1)s^2_B}{n_A +n_B -2}}$

$s_p =\sqrt{\frac{(12 -1)(5100)^2 +(12-1)(5900)^2}{12 +12 -2}}=5514.526$

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.  

With the info given we can replace in formula (1) like this:  

$t=\frac{(37900-39800)-0}{5514.526\sqrt{\frac{1}{12}+\frac{1}{12}}}}=-0.844$  

P value  

We need to find first the degrees of freedom given by:

$df=n_A +n_{B}-2=12+12-2=22$

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

$p_v =2*P(t_{22}$  

Comparing the p value with a significance level for example $\alpha=0.1$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can conclude that the true means are not significantly different.