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Scilla [17]
3 years ago
12

The elasticity of demand for gasoline has been estimated to be 2.0, and the standard error is 1.0. The upper and lower bounds on

the 95 percent confidence interval for the elasticity of demand for gasoline are:
Physics
1 answer:
mr_godi [17]3 years ago
6 0

Answer:

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

\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}   (1)  

Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that z_{\alpha/2}=1.96

The standard error for this case is given:

SE =\frac{\sigma}{\sqrt{n}}=1

Now we have everything in order to replace into formula (1):

2-1.96*11=0.04    

2+1.96*1=3.96  

Explanation:

Previous concepts  

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

\bar X =2 represent the sample mean for the sample  

\mu population mean (variable of interest)  

s represent the sample standard deviation  

n represent the sample size  

Calculate the confidence interval

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

\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}   (1)  

Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that z_{\alpha/2}=1.96

The standard error for this case is given:

SE =\frac{\sigma}{\sqrt{n}}=1

Now we have everything in order to replace into formula (1):

2-1.96*11=0.04    

2+1.96*1=3.96  

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Complete question:

A block of solid lead sits on a flat, level surface. Lead has a density of 1.13 x 104 kg/m3. The mass of the block is 20.0 kg. The amount of surface area of the block in contact with the surface is 2.03*10^-2*m2, What is the average pressure (in Pa) exerted on the surface by the block? Pa

Answer:

The average pressure exerted on the surface by the block is 9655.17 Pa

Explanation:

Given;

density of the lead, ρ =  1.13 x 10⁴ kg/m³

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F = mg

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Determine the pressure exerted on the surface by the block

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