Question

In an effort to make a healthier product, the Oriental Spice Sauce company has reduced the amount of sodium in their product to 900mg. In addition, the standard deviation of the amount of sodium should be 100. To make sure this new product continues to meet the standard, a random sample of 22 bottles is taken, and the standard deviation for the sample was 161.5829. Is there evidence at α=0.025 that the standard deviation of the sodium content exceeds the desired level? Assume the population is normally distributed.

Answer 1

Kindly check explanation

Explanation:

Solution:

Null hypothesis = 100

Alternative hypothesis > 100

Obtain the test statistic:

Same size (n) = 22

Sample standard deviation (s) = 161.5829

Population standard deviation (sd) = 100

Using chisquare :

((n - 1) * s²) / sd²

((22-1) × 161.5829²) / 100²

= (21 × 161.5829) / 10000

= 54.828

Using the online p value calculator from chi square:

Degree of freedom = 22 - 1 = 21

Alpha level = 0.025

The p value = 0.000075

Since the p value obtained is very small, and less than 0.025, and shows that result is significant at p < 0.05/2, then we reject the null hypothesis and conclude that the standard deviation of the sodium content exceeds the desired level.