In an attempt to increase business on Monday nights, a restaurant offers a free dessert with every dinner order. Before the offer, the mean number of dinner customers on Monday was 150. Following are the numbers of diners on a random sample of 12 days while the offer was in effect.

206 169 191 152 212 139 142 151 174 220 192 153Using the critical value method, can you conclude that the mean number of diners increased while the free dessert offer was in effect? Use the alpha = 0.05 level of significance.

Question

Mathematics

Adolfo Compton

1 October, 03:21

In an attempt to increase business on Monday nights, a restaurant offers a free dessert with every dinner order. Before the offer, the mean number of dinner customers on Monday was 150. Following are the numbers of diners on a random sample of 12 days while the offer was in effect.

206 169 191 152 212 139 142 151 174 220 192 153Using the critical value method, can you conclude that the mean number of diners increased while the free dessert offer was in effect? Use the alpha = 0.05 level of significance.

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Rory Crane · Answer 1

Answer: we can conclude that the mean number of diners increased while the free dessert offer was in effect

Step-by-step explanation:

The mean of the set of data given is

Mean = (206 + 169 + 191 + 152 + 212 + 139 + 142 + 151 + 174 + 220 + 192 + 153) / 12 = 175.1

Standard deviation = √ (summation (x - mean) / n

n = 12

Summation (x - mean) = (206 - 175.1) ^2 + (169 - 175.1) ^2 + (191 - 175.1) ^2 + (152 - 175.1) ^2 + (212 - 175.1) ^2 + (139 - 175.1) ^2 + (142 - 175.1) ^2 + (151 - 175.1) ^2 + (174 - 175.1) ^2 + (220 - 175.1) ^2 + (192 - 175.1) ^2 + (153 - 175.1) ^2 = 8910.92

Standard deviation = √ (8910.92/12) = 27.25

We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean

For the null hypothesis,

µ = 150

For the alternative hypothesis,

µ > 150

For the free dessert offer to be in effect, the number of diners would be greater than 150. It means that it is right tailed.

Since the number of samples is 12 and no population standard deviation is given, the distribution is a student's t.

Since n = 12,

Degrees of freedom, df = n - 1 = 11

t = (x - µ) / (s/√n)

Where

x = sample mean = 175.1

µ = population mean = 150

s = samples standard deviation = 27.25

t = (175.1 - 150) / (27.25/√12) = 3.19

We would determine the p value using the t test calculator. It becomes

p = 0.0043

Since alpha, 0.05 > than the p value, 0.0043, then we would reject the null hypothesis. Therefore, At a 5% level of significance, the sample data showed significant evidence that mean number of diners increased while the free dessert offer was in effect