Question

A tire manufacturer is considering a newly designed tread pattern for its all-weather tires. Tests have indicated that these tires will provide better gas mileage and longer tread life. The last remaining test is for braking effectiveness. The company hopes the tire will allow a car traveling at 60 mph to come to a complete stop within an average of 125 feet after the brakes are applied. They will adopt the new tread pattern unless there is strong evidence that the tires do not meet this objective. The distances (in feet) for 10 stops on a test track were 129, 128, 130, 132, 135, 123, 102, 125, 128, and 130. Should the company adopt the new tread pattern? Test an appropriate hypothesis and state your conclusion. Explain how you dealt with the outlier and why you made the recommendation you did.

Answer 1

Answer:

With outlier (102)

$t=\frac{126.2-125}{\frac{9.138}{\sqrt{10}}}=0.415$      

$p_v =P(t_{9}>0.415)=0.344$    

If we compare the p value and a significance level assumed $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true mean is not significantly higher than 125 feet at 5% of significance.    

Without outlier (102)

$t=\frac{128.89-125}{\frac{3.551}{\sqrt{9}}}=3.285$      

$p_v =P(t_{8}>3.285)=0.0056$  

And we conclude that we reject the null hypothesis since $p_v<\alpha$. So the final conclusion would be not use the method since the value of 102 observed can be a potential outlier, removing this value we see that we reject the null hypothesis and we have a significant result that the true mean is higher than 125.

Step-by-step explanation:

Previous concepts  and data given  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

Data: 129, 128, 130, 132, 135, 123, 102, 125, 128, 130

We can calculate the mean with the following formulas:

$\bar X =\frac{\sum_{i=1}^n X_i}{n}$

$s=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

$\bar X=126.2$ represent the sample mean    

$s=9.138$ represent the sample standard deviation  

n=10 represent the sample selected  

$\alpha$ significance level    

State the null and alternative hypotheses.    

We need to conduct a hypothesis in order to check if the mean is significantly higher than 125, the system of hypothesis would be:    

Null hypothesis:$\mu \leq 125$    

Alternative hypothesis:$\mu > 125$    

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:    

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$  (1)    

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".    

Calculate the statistic  

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

$t=\frac{126.2-125}{\frac{9.138}{\sqrt{10}}}=0.415$      

P-value  

First we need to calculate the degrees of freedom given by:

$df=n-1=10-1= 9$

Then since is a right tailed sided test the p value would be:    

$p_v =P(t_{9}>0.415)=0.344$    

Conclusion    

If we compare the p value and a significance level assumed $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true mean is not significantly higher than 125 feet at 5% of significance.    

Using all the data we see that we don;t have enough info to conclude that the true mean is higher than 125, but if we see careful 9 of the 10 values are over the limit of 125 feet. And if we repeat the procedure with the outlier of 102. We got this:

$\bar X=128.89$ represent the sample mean    

$s=3.551$ represent the sample standard deviation

$t=\frac{128.89-125}{\frac{3.551}{\sqrt{9}}}=3.285$      

First we need to calculate the degrees of freedom given by:

$df=n-1=9-1= 8$

Then since is a right tailed sided test the p value would be:    

$p_v =P(t_{8}>3.285)=0.0056$  

And we conclude that we reject the null hypothesis since $p_v<\alpha$. So the final conclusion would be not use the method since the value of 102 observed can be a potential outlier removing this value we see that we reject the null hypothesis and we have a significant result that the true mean is higher than 125.

Answer 2

It should be noted that the test statistic is important in standardizing the mean and one can conclude that the mean is 125.

How to interpret the test statistic

In the information, it can be deduced that the level of significance is 0.05 and it's a two tailed test.

Using x = 126.2, s = 9.14, and n = 10. The value of t will be calculated thus:

t = (126.2 - 125) / (9.14 / ✓10)

= 0.415

Using the degree of freedom of 9, the critical value is 2.262. Therefore, the null hypothesis will not be rejected

At 5% level of significance, the sample evidence shows that the mean is 125.

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