Question

A machine that is programmed to package 2.95 pounds of cereal in each cereal box is being tested for its accuracy. In a sample of 33 cereal boxes, the mean and the standard deviation are calculated as 2.97 pounds and 0.06 pound, respectively. (You may find it useful to reference the appropriate table: z table or table) a. Select the null and the alternative hypotheses to determine if the machine is working improperly, that is, it is either underfilling or overfilling the cereal boxes. How2.95; EN < 2.95 O: N $ 2.95; En » > 2.95 O + x = 2.95; HAN 2.95 b-1. Calculate the value of the test statistic. (Round intermediate calculations to at least 4 decimal places and final answer to 2 decimal places.) Test statistic b-2. Find the p-value 0.02 s p-value < 0.05 0.01 s p value < 0.02 Opvalue <0.01 0.05 s p-value < 0.10 Op value 20:10 9 Next

Answer 1

Given that, sample size (n) = 33, sample mean = 2.97 pounds and sample standard deviation (s) = 0.06 pound

(1)

a) The null and alternative hypotheses are,

H0 : μ = 2.95

HA : μ ≠ 2.95

This hypothesis test is a two-tailed test.

b-1) Test statistic is,

$t = \frac {\bar x - \mu}{\frac {s}{\sqrt n }} = \frac {2.97 - 2.95}{\frac {0.06}{\sqrt {33}}} = 1.91$

=> Test statistic = 1.91

b-2) Degrees of freedom = 33 - 1 = 32

Using t-table we get, p-value for the test statistic = 1.91 with 32 degrees of freedom is between 0.05 and 0.10

Answer : 0.05 ≤ p-value < 0.10

c-1) Do not reject H0, since the p-value is greater than significance level.

c-2) We cannot conclude that the machine is working improperly.

Answer : No

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