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Mashcka [7]
3 years ago
5

company with a large fleet of cars hopes to keep gasoline costs down and sets a goal of attaining a fleet average of at least 26

miles per gallon. To see if the goal is being met, they check the gasoline usage for 50 company trips chosen at random, finding a mean of 25.02 mpg and a standard deviation of 4.83 mpg. Is this strong evidence that they have failed to attain their fuel economy goal
Mathematics
1 answer:
podryga [215]3 years ago
3 0

Answer:

t=\frac{25.02-26}{\frac{4.83}{\sqrt{50}}}=-1.435    

p_v =P(t_{(49)}  

If we compare the p value and the significance level assumed \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the height of men actually its significant lower than 26 so then the specification is satisfied.

Step-by-step explanation:

Data given and notation  

\bar X=25.02 represent the sample mean

s=4.83 represent the sample standard deviation

n=50 sample size  

\mu_o =26 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 check if the mean is at least 26 mpg, the system of hypothesis would be:  

Null hypothesis:\mu \geq 26  

Alternative hypothesis:\mu < 26  

If we analyze the size for the sample is > 30 but 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{25.02-26}{\frac{4.83}{\sqrt{50}}}=-1.435    

P-value

The first step is calculate the degrees of freedom, on this case:  

df=n-1=50-1=49  

Since is a one sided test the p value would be:  

p_v =P(t_{(49)}  

Conclusion  

If we compare the p value and the significance level assumed \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the height of men actually its significant lower than 26 so then the specification is satisfied.

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lorasvet [3.4K]

Answer:

a) Standard Error = 0.010

b) 95% Confidence Interval = (0.0924 , 0.1316)

Step-by-step explanation:

a) The formula for Standard Error = √Sample Proportion (1 - sample proportion)/n

Standard Error = √p (1 - p)/n

We are told in the question that:

In a random sample of 400 people, 112 agree and 288 disagree. Estimate the standard error using 1000 samples

p = x/n

n = 1000 because we were told to use it instead of 400

x = number for people that agree = 112

p = 112/1000

p = 0.112

Standard Error = √p (1 - p)/n

= √0.112 (1 - 0.112)/1000

= √0.112 × 0.888/1000

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= √0.000099456

= 0.0099727629

Approximately to 3 decimal places = 0.010

Therefore, the standard error is 0.010

b) The Question above also asked that we solve for the 95% Confidence Interval

The formula =

p ± z × Standard Error

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z score for 95% confidence interval = 1.96

Standard Error = 0.010

Confidence Interval =

0.112 ± 1.96 × 0.010

= 0.112 ± 0.0196

0.112 - 0.0196

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<em>Explanation:</em>

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In this case, subtracting 3 to all the numbers gave us perfect squares! So this means the nth term has to do with squaring the number and adding three afterward! This can be checked.

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