Question

A variety of packaging solutions exist for products that must be kept within a specific temperature range. Cold chain distribution is particularly useful in the food and pharmaceutical industries. A packaging company is trying out a new packaging material that might reduce the variation of temperatures in the box. It is believed that the temperature in the box follows a normal distribution with both packaging materials. Inspectors randomly select 16 boxes of new and old packages, 24 hours after they are sealed for shipment, and report the temperatures in degrees Celsius. Data is shown in the accompanying table. Assume that the two samples are drawn independently from normally distributed populations. New Package Old Package 4.05 3.76 4.57 4.24 5.93 5.19 6.02 5.39 4.98 5.70 5.88 5.48 3.74 5.83 5.86 3.43 4.75 6.17 3.81 3.59 4.61 5.18 6.58 6.53 6.45 6.16 5.93 6.00 3.77 5.52 3.75 6.83 a.) Let Old Package represent population 1 and New Package represent population 2. Select the appropriate hypotheses to test whether the new packaging material reduces the variation of temperatures in the box. H0: σ12 / σ22 = 1, HA: σ12 / σ22 ≠ 1 H0: σ12 / σ22 ≤ 1, HA: σ12 / σ22 > 1 H0: σ12 / σ22 ≥ 1, HA: σ12 / σ22 < 1 Use Excel's F.TEST function to calculate the p-value.

Answer 1

Solution:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis H₀: σ_New² = σ_Old²

Alternative hypothesis H_A: σ_New² < σ_Old²

Formulate an analysis plan. For this analysis, the significance level is 0.05.

Analyze sample data. Using sample data, the degrees of freedom (DF), and the test statistic (F).

DF₁ = n₁ - 1 = 16 -1

D.F₁ = 15

DF₂ = n₂ - 1 = 16 -1

D.F₂ = 15

Test statistics:-

F = 0.9954

Since the first sample had the smaller standard deviation, this is a left-tailed test.

p value for the F distribution = 0.4965

Interpret results. Since the P-value (0.4965) is greater than the significance level (0.05), we have to accept the null hypothesis.

From the above test there is insufficient evidence to conclude that the the new packaging material reduces the variation of temperatures in the box.