Question

You have 8 collected prices for a textbook (assume that the prices are normally distributed) from various websites and found the mean price to be $214 and sample standard deviation = $26.10. Based on a 90% confidence interval for the population mean, could you can you say that the $200 you paid for a good copy of the book was about the average cost of the textbook or a bargain?

Confidence Interval - _________________________

Conclusion - __________________________________________________

Answer 1

Given that,

= 214

s =26.10

n = 8

Degrees of freedom = df = n - 1 =8 - 1 = 7

At 90% confidence level the t is ,

= 1 - 90% = 1 - 0.90 = 0.1

/ 2 = 0.1 / 2 = 0.05

t _/2,df = t_0.05,7 =1.895

Margin of error = E = t_/2,df * (s /n)

= 1.895* (26.10 / 8) = 17.45

The 90% confidence interval estimate of the population mean is,

- E < < + E

214 - 17.45 < < 214+ 17.45

196.55 < < 231.45

Confidence Interval 196.5 , 231.45

Conclusion

LOWER LIMIT 196.55

UPPER LIMIT 231.45

THAT MEAN VALUE LIES BETWEEN THEN