Question

Quart cartons of milk should contain at least 32 ounces. A sample of 20 cartons contained the following amounts in ounces. Does sufficient evidence exist to conclude the mean amount of milk in cartons is less than 32 ounces at the 5% significance level?
The data is: (32.5, 32.4, 31.8, 28.4, 27.3, 27.2, 28.3, 31.7, 32.8, 31.5, 27.5, 31.8, 28.6, 27.4, 31.7, 32.7, 28.7, 32.1, 32.3, 27.9)

Select the [p-value, Decision to Reject (RH0) or Failure to Reject (FRH0)].

A) [p-value = 0.997, FRH0]
B) [p-value = 0.002, RH0]
C) [p-value = 0.003, RH0]
D) [p-value = 0.003, FRH0]
E) [p-value = 0.997, RH0]

Answer 1

Answer:

B) [p-value = 0.002, RH0]

Step-by-step explanation:

In order to calculate the p-value, we need to find t from the data:

$t=\frac{mean-u}{{\sqrt{\frac{sd^{2} }{n} } } }$

Where, mean is the average of the data set, u is the comparative value (32), sd is the standard deviation and n is the number of elements.

And the standard deviation is:

∑$\sqrt{\frac{(x-mean)^{2} }{n-1} }$

$mean=\frac{604.6}{20}$

mean=30.23

sd=2.2

$t=\frac{30.23-32}{{\sqrt{\frac{2.2^{2} }{20} } } }$

t=-3.59

From the t-table with degrees of freedom= df = n-1=19 we get a p-value of 0.002

We a significance level of 5%, or α=0.05

So we have two options:

• If p-value ≤α RHo
• If p-value >α FRHo

In this case 0.002<0.05 or option B.