2 years ago

VI. PRACTICE TASKS

Mathematics

1 answer:

dimulka [17.4K]2 years ago

8 0

Step-by-step explanation:

1 _ 1.215

2_ 4.0608

3_1.5225

4_0.2875

5_3.6064

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In analyzing the city and highway fuel efficiencies of many cars and trucks, the mean difference in fuel efficiencies for the 60

Olegator [25]

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$7.38-1.96\frac{2.51}{\sqrt{601}}=7.18$

$7.38+1.96\frac{2.51}{\sqrt{601}}=7.58$

For this case the confidence interval is given by :

$7.18 \leq \mu \leq 7.58$

Step-by-step explanation:

Data provided

$\bar X=7.38$ represent the sample mean for the fuel efficiencies

$\mu$ population mean

s=2.51 represent the sample standard deviation

n=601 represent the sample size

Confidence interval

The formula for the confidence interval of the true mean is given by:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

The degrees of freedom for this case is given by:

$df=n-1=601-1=600$

The Confidence level for this case is 0.95 or 95%, and the significance level $\alpha=0.05$ and $\alpha/2 =0.025$ and the critical value is given by $t_{\alpha/2}=1.96$

Replcing into the formula for the confidence interval is given by:

$7.38-1.96\frac{2.51}{\sqrt{601}}=7.18$

$7.38+1.96\frac{2.51}{\sqrt{601}}=7.58$

For this case the confidence interval is given by :

$7.18 \leq \mu \leq 7.58$

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