Question

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 Mondays while the offer was in effect.

181,172,1 53, 142, 160

156​,

144​,

152​,

176​,

172

Can you conclude that the mean number of diners increased while the offer was in effect? Use the α = 0.05 level of significance and assume the population is normally distributed. Use the p-value method showing the RANGE that the p-value should fall in. Do NOT give the EXACT p-value.

Answer 1

∑x = 1608

∑x² = 260254

n = 10

Mean , x̅ = Ʃx/n = 1608/10 = 160.8

Standard deviation, s = √[(Ʃx² - (Ʃx)²/n)/(n-1)] = √[(260254-(1608)²/10)/(10-1)] = 13.6935

Null and Alternative hypothesis:

Ho : µ = 150

H1 : µ > 150

Test statistic:

t = (x̅- µ)/(s/√n) = (160.8 - 150)/(13.6935/√10) = 2.4941

df = n-1 = 9

p-value = T.DIST.RT(2.4941, 9) = 0.0171

Range: 0.01 < p-value < 0.025

Decision:

p-value < α, Reject the null hypothesis

There is enough evidence to conclude that the mean number of diners increased while the offer was in effect at 0.05 significance level.