Question

a restaurant owner is interested in figuring out how to increase the average restaurant bill per guest. after changing the menu by adding more items, a random sample of 150 guests shows that the average bill per guest is $28.60. he believes that the strategy of changing the menu affected the average bill per guest. assume that the standard deviation of the population is known and it is 4.70. at the 5% significance level, how sure can he be that the average bill per guest is not $28 (the previous average bill per guest)? in other words, is there sufficient evidence to reject the null hypothesis? and your response to the question. (provide your answer with two decimal places, rounding to the nearest hundredths, as in x.xx) First, what is the p-value?(Provide your answer with two decimal places, rounding to the nearest hundredths, as in x.xx)
Second, at the 5% significance level, how sure can he be that the average bill per guest is not $28 (the previous average bill per guest)? In other words, is there sufficient evidence to reject the null hypothesis? Explain your answer.

Answer 1

The confidence interval for the given hypothesis of a restaurant owner to increase the average restaurant bill per guest be, (144.83 , 155.17) and the margin of error of the poll be, ±5.17

On Subtracting 1 from your sample size, we get

150 – 1 = 149.

On Subtracting confidence level from 1, and then divide by two.

(1 – .95) / 2 = 0.025

Now, df = 149 and α = 0.025

from the table at df = 149 we got 2.262.

Divide your sample standard deviation by the square root of your sample size.

28 / √(150) = 2.28

Now, 2.262 × 2.28 = 5.17

So, the confidence interval be,

(150 - 5.17 , 150 + 5.17) = (144.83 , 155.17)

Hence, the confidence interval for the given hypothesis of a restaurant owner to increase the average restaurant bill per guest be, (144.83 , 155.17) and the margin of error of the poll be, ±5.17

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