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vlada-n [284]
3 years ago
11

According to the Environmental Protection Agency, the mean miles per gallon of all large cars manufactured in 2012 driven under

highway driving conditions is 25.1 mpg. A researcher claims that a new fuel additive will increase the miles per gallon of cars. After obtaining a random sample of 35 large cars from 2012, the researcher adds the fuel additive. The sample mean is found to be 26.8 mpg, with a standard deviation of 3.9. Test the researcher's claim at the .05 level of significance.
Mathematics
1 answer:
Whitepunk [10]3 years ago
8 0
Step 1: Writing Null and Alternate Hypothesis

First we need to write the Null and Alternate Hypothesis for this problem.

Researcher claims that mileage will be increased on addition of Fuel additive. So the null and alternate hypothesis will be:

H_{o}: μ ≤ 25.1        (Null Hypothesis)
H_{a}: μ > 25.1        (Alternate Hypothesis)

This is a Right Tailed Test. Since population standard deviation is not know, we will use t-test to check the researchers claim.

Step 2: Finding Test Statistic

Sample Mean = x = 26.8
Standard Deviation = s = 3.9
Sample Size = n = 35 
Degrees of Freedom = df = n - 1 = 34

Test statistic(t) is given by:

t= \frac{x-u}{ \frac{s}{ \sqrt{n} } } \\  \\ 
t= \frac{26.8-25.1}{ \frac{3.9}{ \sqrt{35} } } \\  \\ 
t=2.579

Step 3: Finding p value

Using t table or calculators find the p value for t=2.579 with 34 degrees of freedom for one tailed test.

P value comes out to be:
p = 0.0072

Step 4: Conclusion

Since the p value is lesser than the significance level of 0.05, we reject the Null Hypothesis.

We have enough evidence to support the researcher's claim that the new additive increases the miles per gallon of the cars. 
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Rosetta wants to estimate the percentage of people who rent their home. She surveys 250 individuals and finds that 48 rent their
Andreas93 [3]

Answer:

0.192 - 1.645\sqrt{\frac{0.192(1-0.192)}{250}}=0.151

0.192 + 1.645\sqrt{\frac{0.192(1-0.192)}{250}}=0.233

Step-by-step explanation:

Information given

X= 48 number of people who rent their home

n= 250 represent the sample size

\hat p =\frac{48}{250}= 0.192 represent the proportion of people who rent their home

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 90% of confidence, our significance level would be given by \alpha=1-0.90=0.1 and \alpha/2 =0.05. And the critical value would be given by:

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The confidence interval for the mean is given by the following formula:  

\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}

If we replace the values obtained we got:

0.192 - 1.645\sqrt{\frac{0.192(1-0.192)}{250}}=0.151

0.192 + 1.645\sqrt{\frac{0.192(1-0.192)}{250}}=0.233

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