Some manufacturers claim that non-hybrid sedan cars have a lower mean miles-per-gallon (mpg) than hybrid ones. Suppose that consumers test  256
  hybrid sedans and get a mean of  28
  mpg with a standard deviation of  4.25
  mpg. Also,  246
  non-hybrid sedans get a mean of  27
  mpg with a standard deviation of  6.5
  mpg. Suppose that both populations are known to be normal distributed.

Conduct a hypothesis test to evaluate the manufacturers' claim. Test at a  7%
  level of significance.



Let  population  1
  denote non-hybrids and population  2
  denote hybrids.

 

(a)  H0
 :  μ1
   
   μ2
 
 

 

(b)  Ha
 :  μ1
   
   μ2
 
 

 

(c) In words, state what your random variable represents.

The random variable is the  
  miles per gallon of non-hybrid sedans and hybrid sedans.

 

(d) Select the distribution to use for the test.

	
Xˉ1−Xˉ2∼N[μ1−μ2,σ]
 

	
Xˉ1−Xˉ2∼N[μ,σ]
 

	
tdf
 

	
Xˉ∼N[μ,σ]
 

	
Xˉ1−Xˉ2∼N[σ,μ]
 

 

(e) What is the test statistic?

Round your answer to three decimal places


Let's go through each part of the problem step by step.

### (a) Null Hypothesis (\(H_0\)):
The null hypothesis is that the mean miles-per-gallon (mpg) of non-hybrid sedans (\(\mu_1\)) is greater than or equal to the mean mpg of hybrid sedans (\(\mu_2\)).

\[
H_0: \mu_1 \geq \mu_2
\]

### (b) Alternative Hypothesis (\(H_a\)):
The alternative hypothesis is that the mean mpg of non-hybrid sedans (\(\mu_1\)) is less than the mean mpg of hybrid sedans (\(\mu_2\)).

\[
H_a: \mu_1 < \mu_2
\]

### (c) Random Variable:
The random variable represents the difference in miles-per-gallon between non-hybrid sedans and hybrid sedans.

### (d) Distribution to Use for the Test:
Given that the sample sizes are large (\(n_1 = 246\) and \(n_2 = 256\)), the sampling distribution of the difference in sample means (\(\bar{X}_1 - \bar{X}_2\)) can be approximated by a normal distribution.

\[
\bar{X}_1 - \bar{X}_2 \sim N\left(\mu_1 - \mu_2, \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}\right)
\]

### (e) Test Statistic:
The test statistic for comparing two means is given by:

\[
z = \frac{\bar{X}_1 - \bar{X}_2 - 0}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}}
\]

Where:
- \(\bar{X}_1 = 27\) (mean of non-hybrid mpg)
- \(\bar{X}_2 = 28\) (mean of hybrid mpg)
- \(\sigma_1 = 6.5\) (standard deviation of non-hybrid mpg)
- \(\sigma_2 = 4.25\) (standard deviation of hybrid mpg)
- \(n_1 = 246\) (sample size of non-hybrids)
- \(n_2 = 256\) (sample size of hybrids)

Now, let's calculate the test statistic:

\[
z = \frac{27 - 28}{\sqrt{\frac{6.5^2}{246} + \frac{4.25^2}{256}}}
\]

I'll calculate this value.The test statistic \(z\) is approximately \(-2.032\).

Would you like to proceed with the next steps, such as finding the p-value or making a decision based on the test statistic? 

Here are 5 related questions to consider:
1. How do you interpret the p-value in the context of this hypothesis test?
2. What is the critical value for a one-tailed test at the 7% significance level?
3. How does the large sample size affect the approximation to the normal distribution?
4. What assumptions must be met for the test statistic to be valid?
5. How would the conclusion change if the significance level was set at 5% instead of 7%?

**Tip:** Always remember to check the assumptions of normality and equal variances before conducting a hypothesis test for means.

Conduct a hypothesis test to evaluate whether non-hybrid sedan cars have a lower mean miles-per-gallon (mpg) than hybrid sedans. Analyze sample data with specific means and standard deviations using a z-test. Suitable for advanced high school students studying statistics and hypothesis testing.

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