Question

A manufacturer of banana chips would like to know whether its bag filling machine works correctly at the 432 gram setting. It is believed that the machine is underfilling the bags. A 24 bag sample had a mean of 426 grams with a variance of 225. A level of significance of 0.01 will be used. Assume the population distribution is approximately normal. Determine the decision rule for rejecting the null hypothesis. Round your answer to three decimal places.

Answer 1

Answer:

The decision rule will be as follows: Reject the null hypothesis if test-statistic < -2.5 or fail to reject the

Now, Our test-statistic was -1.96, which is greater than -2.5. Thus, we will fail to reject the null hypothesis.

Step-by-step explanation:

We are given;

Population mean; μ = 432

Sample mean; x¯ = 426

Sample size; n = 24

Variance; Var = 225

Now, standard deviation from variance is; σ = √Var

Thus, σ = √225

σ = 15

Hypothesis is defined as;

Null hypothesis; H0: μ ≥ 432

Alternative hypothesis: Ha: μ < 432

Formula for the test statistic since sample size is less than 30 is;

t = (x¯ - μ)/(σ/√n)

Plugging in the relevant values;

t = (426 - 432)/(15/√24)

t = -1.96

Now,we are given significance level of 0.01; our DF = n - 1 = 24 - 1 = 23.

From t-table attached, we can see that at DF of 23 and significance value of 0.01, the critical value is 2.5

However, this is a left tailed test as the rejection region is to the left.

Thus, the critical value of a left tailed test is negative. Thus, in this case our critical value will be -2.5.

The decision rule will be as follows: Reject the null hypothesis if test-statistic < -2.5 or fail to reject the

Now, Our test-statistic was -1.96, which is greater than -2.5. Thus, we will fail to reject the null hypothesis.