A restaurant tried to increase business on Monday night, by featuring a special $1.00 dessert menu....

A restaurant tried to increase business on Monday night, by featuring a special $1.00 dessert menu. The number of diners on each of 11 Mondays recorded while the special menu was in effect.

The data were

119 139 121 126 128 108 63 118 109 131 142

a. Find a 95% confidence interval the long-run number of diners.

b. Before the special menu, the restaurant average 105.2 diners per Monday night. Is it t reasonable to interpret the confidence interval from part “a” as indicating that the special menu did not increase the average number of diners?

Expert Solution

a. Let us first find mean and standard deviation of sample data

Create the following table.

data	data-mean	(data - mean)2
119	0.4545	0.20657025
139	20.4545	418.38657025
121	2.4545	6.02457025
126	7.4545	55.56957025
128	9.4545	89.38757025
108	-10.5455	111.20757025
63	-55.5455	3085.30257025
118	-0.5455	0.29757025
109	-9.5455	91.11657025
131	12.4545	155.11457025
142	23.4545	550.11357025

Find the sum of numbers in the last column to get.

So sd is

t value for 10 df is TINV(0.05,10)=2.228

So Margin of Error is

CI is

b. As 105.2 lies in the range, it is reasonable to interpret the confidence interval from part “a” as indicating that the special menu did not increase the average number of diners

orchestra answered 1 year ago

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