Question

A manufacturer of chocolate chips would like to know whether its bag filling machine works correctly at the 420.0 gram setting. It is believed that the machine is underfilling the bags. A 33 bag sample had a mean of 417.0 grams. A level of significance of 0.01 will be used. Determine the decision rule. Assume the variance is known to be 576.00.

Answer 1

The decision rule is

Fail to reject the null hypothesis

The conclusion is

There is no sufficient evidence to conclude that the machine is underfilling the bags

Step-by-step-explanation:

From the question we are told that

The population mean is
`\mu = 420.0 \ g`

The sample size is n = 33 bags

The sample mean is
`\= x = 417.0 \ g`

The level of significance is
`\alpha = 0.01`

The variance is
`\sigma^2 = 576.0`

The null hypothesis is
`H_o : \mu = 420`

The alternative hypothesis is
`H_a : \mu < 420`

Generally the standard deviation is mathematically represented as

`\sigma = √(\sigma^2)`

=>
`\sigma = √(576 )`

=>
`\sigma = 24`

Generally the test statistics is mathematically represented as

`z = ( \= x - \sigma )/( (\sigma)/(√(n) ) )`

=>
`z = (417 - 420 )/( (24)/( √( 33) ) )`

=>
`z = -0.71807`

From the z table the area under the normal curve to the left corresponding to -0.71807 is

`p-value = P(Z < -0.71807 ) = 0.23636`

Generally looking at the values we see that

`p-value > \alpha` , hence

The decision rule is

Fail to reject the null hypothesis

The conclusion is

There is no sufficient evidence to conclude that the machine is underfilling the bags