Question

A manufacturer of salad dressings uses machines to dispense liquid ingredients into bottles that move along a filling line. The machine that dispenses salad dressings is working properly when 8 ounces are dispensed. Suppose that the average amount dispensed in a particular sample of 35 bottles is 7.91 ounces with a variance of 0.03 ounces squared. Is there evidence that the machine should be stopped and production wait for repairs? The lost production from a shutdown is potentially so great that management feels that the level of significance in the analysis should be 99%.

Answer 1

Answer:

Step-by-step explanation:

From the given information:

The null hypothesis is:

$H_o: \mu = 8$

The alternative hypothesis is:

$H_1 : \mu \ne 8$

The sample size n = 35

Sample mean = 7.91

variance = 0.03

Standard deviation = $\sqrt{0.03} = 0.173$

The t-test statistics is computed as:

$t = \dfrac{\bar x - \mu}{\dfrac{s}{\sqrt{n}}}$

$t = \dfrac{7.91 -8}{\dfrac{0.1732}{\sqrt{35}}}$

$t = \dfrac{-0.09}{\dfrac{0.1732}{5.916}}$

t = - 3.0741

Degree of freedom:

df = n - 1

df = 35 - 1

df = 34

Level of significance = 1 - C.I

= 1 - 0.99

= 0.01

The t_{critical} value for the two-tailed test at 0.01 is within the rejection region:

$t_{critical} = -2.728 \ and \ 2.728$

Decision rule: To reject $H_o$ if  $t< -2.728 \ or \ t > 2.728$

As $t_{statistics}$ fails the rejection region, we reject $H_o$

Conclusion: There is sufficient evidence to believe that the machine will not dispense 8 ounces & and it must be stopped for repairs.