As quality control manager at a raisin manufacturing and packaging plant, you want to ensure that all the boxes of raisins you sell are comparable, with 30 raisins in each box (assume the population average number of raisins per box is 30). In the plant, raisins are poured into boxes until the box reaches its sale weight. To determine whether a similar number of raisins are poured into each box, you randomly sample 36 boxes about to leave the plant and count the number of raisins in each. In this sample, you find the mean number of raisins in each box to be 28.05, with a standard deviation of 3.87.
1. Perform a two-tailed, single sample t test (p<.05) to determine whether the average number of raisins per box differs from the expected 30.
2. What type of statistical analysis should be run to address this question?3. What are the assumptions of your chosen statistical test?4. Identify the null and alternative/research hypotheses in both words and symbols.

The type of statistical analysis to be run here is the inferential statistical analysis and its assumptions include normality, linearity and equality of variance.

Null hypothesis: Average number of raisins per box is 30: u = 30

Alternative hypothesis: Average number of raisins per box is not 30: u ≠ 30

To perform a two tailed single sample test, we have

z statistic = (the sample mean - hypothesized mean) / (SD/√n)

sample mean = 28.05, SD = 3.87 hypothesized mean = 30 n = 36

Thus, we have 28.05 - 30 / (3.87/√36)

= -1.95 / (3.87/6)

= -1.95 / 0.645

= -3.023

we say this is a left tailed test and use the negative z table

where -3.02 = 0.0013

To determine whether the average number of raisins per box differs from the expected 30.

Using p < 0.05, since the observed value 0.0013 is less than 0.05, we conclude the the average number of raisins per box differs from the expected 30.