Question

Problem 3 (30%) A local pizzeria offers a “Pepperoni Feast" pizza advertised to be covered with (on average) 100 slices of pepperoni. You've had some time on your hands lately, so you've counted the number of slices of pepperoni on each of the nine Pepperoni Feast pizzas you've ordered in the past few months. You've collected the following series of results: 97, 105, 90, 92, 89, 92, 97, 105, 94. Based on your data, can the pizzeria's claim that Pepperoni Feast pizzas contain an average of 100 slices of pepperoni be disproved at the (a, 15%) 95% confidence level? (b, 15%) 99% confidence level?

Answer 1

a) 95% confidence level

$\dpi{100} \alpha =0.05$

The null and alternative hypothesis

H0: $\dpi{100} \mu =100$

Ha : $\dpi{100} \mu <100$

Test statistic

$\dpi{100} t= \frac{\bar{X}-\mu }{s /\sqrt{n}}$

n=9 ,

$\dpi{100} \bar{x} = \frac{\sum x}{n}= 861/9= 95.67$

$\dpi{100} s= \sqrt{\sum (x-\bar{x})^{2}/(n-1)}$

= $\dpi{100} \sqrt{284/(9-1)}=5.96$

x	(x-xbar)^2
97	1.7689
105	87.0489
90	32.1489
92	13.4689
89	44.4889
92	13.4689
97	1.7689
105	87.0489
94	2.7889
861	284.0001

Therefore

$\dpi{100} t=\frac{95.67-100}{5.96/3}$

= -3.07

degrees of freedom = 9-1 =8

P value = 0.0076 ( one tailed )

Since P value < 0.05

We reject H0

There is sufficient evidence to conclude at 95% confidence that pizzaria's claim of average 100 pepperoni slices can be disproved .

Note : P value from excel " =T.DIST.RT(3.07,8)"

b) 99% confidence

$\dpi{100} \alpha =0.01$

P value =0.0076 < 0.01

We reject H0

There is sufficient evidence to conclude at 99% confidence that pizzaria's claim of average 100 pepperoni slices can be disproved .