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Xelga [282]
3 years ago
13

The research department at the company took a sample of 25 comparable textbooks and collected information on their prices. This

information produces a mean price of $145 for this sample. It is known that the standard deviation of the prices of all such textbooks is $35 and the population of such prices is normal. (a) What is the point estimate of the mean price of all such textbooks? (b) Construct a 90% confidence interval for the mean price of all such college textbooks.
Mathematics
1 answer:
Katarina [22]3 years ago
3 0

Answer:

a) \hat \mu = \bar X = 145

b) 145-1.64\frac{35}{\sqrt{25}}=133.52  

145+1.64\frac{35}{\sqrt{25}}=156.48  

So on this case the 90% confidence interval would be given by (133.52;156.48)  

Step-by-step 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=145 represent the sample mean  

\mu population mean (variable of interest)  

\sigma=35 represent the population standard deviation  

n=25 represent the sample size  

a) For this case the best point of estimate for the population mean is the sample mean:

\hat \mu = \bar X = 145

b) 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 0.90 or 90%, the value of \alpha=0.1 and \alpha/2 =0.05, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.05,0,1)".And we see that z_{\alpha/2}=1.64

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

145-1.64\frac{35}{\sqrt{25}}=133.52  

145+1.64\frac{35}{\sqrt{25}}=156.48  

So on this case the 90% confidence interval would be given by (133.52;156.48)  

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