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Mnenie [13.5K]
1 year ago
7

Suppose that the Department of Transportation would like to test the hypothesis that the average age of cars on the road is less

than 12 years. A random sample of 45 cars had an average age of 10.6 years. It is believed that the population standard deviation for the age of cars is 4.1 years. The Department of Transportation would like to set 0.05. What is the conclusion for this hypothesis test? A. Since the test statistie is more than the critical value, we can condude that the average age of cars on the road is less than 12 years. B. Since the test statistic is less than the critical value, we can conclude that the average age of cars on the road is less than 12 years. c. Since the test statistic is less than the critical value, we cannot conclude that the average age of cars on the road is less than 12 years. D. Since the test statistic is more than the critical value, we cannot conclude that the average age of cars on the road is less than 12 years.
Mathematics
1 answer:
dimulka [17.4K]1 year ago
3 0

Answer:

z=\frac{10.6-12}{\frac{4.1}{\sqrt{45}}}=-2.29  

z_{critc}= -1.64

Since our calculates values is lower than the critical value we have enough evidence to reject the null hypothesis at 5% of significance. And the best conclusion for this case is:

B. Since the test statistic is less than the critical value, we can conclude that the average age of cars on the road is less than 12 years.

Step-by-step explanation:

Data given and notation  

\bar X=10.6 represent the sample mean

\sigma=4.1 represent the population standard deviation

n=45 sample size  

\mu_o =12 represent the value that we want to test  

\alpha=0.05 represent the significance level for the hypothesis test.  

z 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 check if the mean is less than 12, the system of hypothesis would be:  

Null hypothesis:\mu \geq 12  

Alternative hypothesis:\mu < 12  

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}} (1)  

z-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:  

z=\frac{10.6-12}{\frac{4.1}{\sqrt{45}}}=-2.29  

Critical value

For this case since w ehave a left tailed distribution we need to find a value who accumulates 0.05 of the area on th left in the normal standard distribution, and we can use the following excel code:

"=NORM.INV(0.05,0,1)"

z_{critc}= -1.64

Conclusion  

Since our calculates values is lower than the critical value we have enough evidence to reject the null hypothesis at 5% of significance. And the best conclusion for this case is:

B. Since the test statistic is less than the critical value, we can conclude that the average age of cars on the road is less than 12 years.

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Answer:

One can be 99% confident the true mean shell length lies within the above interval.

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Step-by-step explanation:

We are given that Time-depth recorders were deployed on 6 of the 76 captured turtles. These 6 turtles had a mean shell length of 51.3 cm and a standard deviation of 6.6 cm.  

The pivotal quantity for a 99% confidence interval for the true mean shell length is given by;

                    P.Q.  =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

where, \bar X = sample mean shell length = 51.3 cm

             s = sample standard deviation = 6.6 cm

             n = sample of turtles = 6

             \mu = true mean shell length

Now, the 99% confidence interval for \mu =  \bar X \pm t_(_\frac{\alpha}{2}_)  \times \frac{s}{\sqrt{n} }

Here, \alpha = 1% so  (\frac{\alpha}{2}) = 0.5%. So, the critical value of t at 0.5% significance level and 5 (6-1) degree of freedom is 4.032.

<u>So, 99% confidence interval for</u> \mu  =  51.3 \pm 4.032 \times \frac{6.6}{\sqrt{6} }

                                                         = [51.3 - 10.864 , 51.3 + 10.864]

                                                         = [40.44 cm, 62.16 cm]

The interpretation of the above result is that we are 99% confident that the true mean shell length lie within the above interval of [40.44 cm, 62.16 cm].

The assumption about the distribution of shell lengths must be true in order for the confidence interval, part a, to be valid is that;

C. The population has a relative frequency distribution that is approximately normal.

This assumption is reasonably satisfied as the data comes from the whole 76 turtles and also we don't know about population standard deviation.

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