Local Pizzeria A local pizzeria wants to know how much the a

Local Pizzeria. A local pizzeria wants to know how much the average cost of an order is during lunchtime on Wednesdays. They collected data from ten orders on Wednesdays (assume their sampling method provides a representative sample), and the cost of an order during lunchtime on Wednesdays is approximately normally distributed. The data from their sample was: $6.18 $11.98 $13.55 $18.99 $20.54 $23.35 $35.00 $37.33 $40.77 $46.71 (g) [Free Response] Interpret the confidence interval you constructed above in the context of the problem. (h) Assume that the true standard deviation of the cost of lunchtime orders on Wednesdays is $13.677 (sigma = 13.677) .i.e, sigma is known. Calculate your 95% confidence interval under this situation. What is the value of the lower bound (lower end point) of the 95% confidence interval? (Round your answer to 2 decimal places) (i) Assume that the true standard deviation of the cost of lunchtime orders on Wednesdays is $13.677 (a = 13.677) .i.e, u is known. Calculate your 95% confidence interval tinder this situation. What is the value of the tipper bound (upper end point) of the 95% confidence interval? (Round your answer to 2 decimal places) (j) [Free Response] Is your 95% confidence interval wider or narrower in this case (part (h) and (i)) than the interval you constructed before? Explain.

Solution

Mean = 25.46

std dev = 12.973

std error = 12.973/rt 10 = 4.102

95% confidence interval margin of error = 1.96(4.102) = 8.04

Confidence interval = (17.42,33.50)

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If sigma = 13.677 is given std error = 4.325

Margin of error = 8.477

Conf interval = (16.98,33,93)

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This is wider than our previous part. This is because std dev is higher.

When std dev increases, sigma increases as a result margin of error increases widening the confidence interval.