Question

A manufacturer of banana chips would like to know whether its bag filling machine works correctly at the 433 gram setting. It is believed that the machine is underfilling the bags. A 15 bag sample had a mean of 423 grams with a standard deviation of 26. A level of significance of 0.1 will be used. Assume the population distribution is approximately normal. Determine the decision rule for rejecting the null hypothesis.

Answer 1

`t=(423-433)/((2.6)/(√(15)))=-14.896`

`df=n-1=15-1=14`

We need to find in the t distribution with df=14 a value who accumulates 0.1 of the area in the left and we got
`t_(crit)= -1.345`.

Since our calculated value for the statistic is is so much lower than the critical value we have enough evidence to reject the null hypothesis, and we can conclude that the true mean for this case is significantly less than 433 and then the machine is underfilling.

Explanation:

Data given

`\bar X=423` represent the sample mean

`s=26` represent the sample standard deviation

`n=15` sample size

`\mu_o =433` represent the value that we want to test

`\alpha=0.1` represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

`p_v` represent the p value for the test

System of hypothesis

We need to conduct a hypothesis in order to check if the true mean is less than 433 (underfilling), the system of hypothesis would be:

Null hypothesis:
`\mu \geq 433`

Alternative hypothesis:
`\mu < 433`

The statistic is:

`t=(\bar X-\mu_o)/((s)/(√(n)))` (1)

Calculate the statistic

`t=(423-433)/((2.6)/(√(15)))=-14.896`

Decision rule

The degrees of freedom are:

`df=n-1=15-1=14`

We need to find in the t distribution with df=14 a value who accumulates 0.1 of the area in the left and we got
`t_(crit)= -1.345`

Since our calculated value for the statistic is is so much lower than the critical value we have enough evidence to reject the null hypothesis, and we can conclude that the true mean for this case is significantly less than 433 and then the machine is underfilling.