Question

A car manufacturer is trying to develop a new sports car. Engineers are hoping that the average amount of time that the car takes to go from 0 to 60 miles per hour is below 6 seconds. The car company tested 12 of the cars and clocked their performance times. Three of the cars clocked in at 5.8 seconds, 5 cars at 5.9 seconds, 3 cars at 6.0 seconds, and 1 car at 6.1 seconds. At a 5% level of significance, test if the new sports car is meeting its goal to go from 0 to 60 miles per hour in less than 6 seconds. Assume a normal distribution for the analysis.a. Specify the competing hypotheses to test this belief.a. H0: μ = 6; HA: μ ≠ 6b. H0: μ ≥ 6; HA: μ < 6c. H0: μ ≤ 6; HA: μ > 6b-1. Calculate the value of the test statistic. (Negative value should be indicated by a minus sign. Round your answer to 2 decimal places.)
b-2. Select the p-value.
a. 0.005 < p-value < 0.01
b. p-value < 0.01
c. 0.01 ≤ p-value < 0.025
d. 0.025 ≤ p-value < 0.05
e. 0.05 ≤ p-value < 0.10
d. p-value ≥ 0.10

Answer 1

Answer:

b. H0: μ ≥ 6; HA: μ < 6

$t=\frac{5.917-6}{\frac{0.0937}{\sqrt{12}}}=-3.07$        

$df = n-1 = 12-1 =11$

$p_v =P(t_{11}<-3.069)=0.0053$    

Step-by-step explanation:

Data given and notation        

Data: 5.8,5.8,5.8,5.9,5.9,5.9,5.9,5.9, 6,6,6,6.1

We can calculate the sample mean and deviation with the following formulas:

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

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

$\bar X=5.917$ represent the mean for the sample    

$s=0.0937$ represent the standard deviation for the sample

$n=12$ sample size        

$\mu_o =6$ represent the value that we want to test      

$\alpha$ represent the significance level for the hypothesis test.      

t would represent the statistic (variable of interest)        

$p_v$ represent the p value for the test (variable of interest)    

State the null and alternative hypotheses.        

We need to conduct a hypothesis in order to determine if the true mean is lower than 6:      

Null hypothesis:$\mu \geq 6$        

Alternative hypothesis:$\mu < 6$        

We don't know the population deviation, so for this case 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:        

$t=\frac{5.917-6}{\frac{0.0937}{\sqrt{12}}}=-3.07$        

Calculate the critical value

The degrees of freedom are given by:

$df = n-1 = 12-1 =11$

The critical value for this case would be :

$P(t_{11}<a)=0.05$

The value of a that satisfy this on the t distribution with 11 degrees of freedom is a=-1.796 and would be the critical value on this case zc=-1.796.

Calculate the P-value        

Since is a one-side lower test the p value would be:        

$p_v =P(t_{11}<-3.069)=0.0053$    

Conclusion        

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v<\alpha$ so we can conclude that we reject the null hypothesis, and the the actual mean is significantly lower than 6

Answer 2

Using the t-distribution, we have that:

a) H0: μ ≥ 6; HA: μ < 6

b) The test statistic is t = -3.08

c) a. 0.005 < p-value < 0.01

Item a:

At the null hypothesis, it is tested if the mean is of at least 6 seconds, that is:

$H_0: \mu \geq 6$

At the alternative hypothesis, it is tested if the mean is of less than 6 seconds, that is:

$H_1: \mu < 6$

Item b:

We have the standard deviation for the sample, thus, the t-distribution is used. The test statistic is given by:

$t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}$

The parameters are:

• $\overline{x}$ is the sample mean.
• $\mu$ is the value tested at the null hypothesis.
• s is the standard deviation of the sample.
• n is the sample size.

For this problem, the values of the parameters are:

$\mu = 6, n = 12$

$\overline{x} = \frac{3(5.8) + 5(5.9) + 3(6) + 6.1}{12} = 5.92$

$s = \sqrt{\frac{3(5.8-5.92)^2 + 5(5.9-5.92)^2 + 3(6-5.92)^2 + (6.1-5.92)^2}{12}} = 0.09$

Hence, the value of the test statistic is:

$t = \frac{5.92 - 6}{\frac{0.09}{\sqrt{12}}}$

$t = -3.08$

Item c:

The p-value of the test is found using a left-tailed test, as we are testing if the mean is less than a value, with t = -3.08 and 12 - 1 = 11 df.

Using a t-distribution calculator, it is of 0.0052, hence:

a. 0.005 < p-value < 0.01

A similar problem is given at https://brainly.com/question/13873630