Question

The quality control department of a soup brand is concerned that one of its soups contains less sodium than the 747 mg per serving listed on the can. A sample of 26 cans from the canning facility had an average sodium of 751 mg and a variance of 3364.00 mg per serving. What can be concluded with an α of 0.05?

a) What is the appropriate test statistic?
---Select--- na z-test One-Sample t-test Independent-Samples t-test Related-Samples t-test
Population:
---Select--- sodium soup brand quality control department sodium listed on can cans from facility
Sample:
---Select--- sodium soup brand quality control department sodium listed on can cans from facility

b) Compute the appropriate test statistic(s) to make a decision about H₀.
(Hint: Make sure to write down the null and alternative hypotheses to help solve the problem.)
critical value =  ; test statistic =  
Decision:  ---Select--- Reject H0 Fail to reject H0

c) If appropriate, compute the CI. If not appropriate, input "na" for both spaces below.
[  ,  ]
) Compute the corresponding effect size(s) and indicate magnitude(s).
If not appropriate, input and/or select "na" below.
d =  ;   ---Select--- na trivial effect small effect medium effect large effect
r² =  ;   ---Select--- na trivial effect small effect medium effect large effect

d) Make an interpretation based on the results.

The amount of sodium in the soup is significantly more than the general listing on the can.The amount of sodium in the soup is significantly less than the general listing on the can.    The amount of sodium in the soup is not significantly different the general listing on the can.

Answer 1

a) What is the appropriate test statistic?

Answer: One-Sample t-test

Population: All cans from facility

Sample: 26 cans from facility

b) Compute the appropriate test statistic(s) to make a decision about H₀.

We are given

Level of significance = α = 0.05

Sample size = n = 26

Degrees of freedom = n – 1 = 25

So, critical value = -1.7081

(by using t-table)

Test statistic is given as below:

t = (Xbar - µ)/[S/sqrt(n)]

We are given

Xbar = 751

S^2 = 3364

S = sqrt(3364) = 58

µ = 747

n = 26

t = (751 – 747)/[58/sqrt(26)]

t = 0.3517

Test statistic = 0.3517

Test statistic value is greater than critical value, so we fail to reject H₀

c) If appropriate, compute the CI. If not appropriate, input "na" for both spaces below.

Confidence interval for Population mean is given as below:

Confidence interval = Xbar ± t*S/sqrt(n)

Confidence level = 95%

df = 25

Critical t value = 2.0595

(by using t-table)

Confidence interval = Xbar ± t*S/sqrt(n)

Confidence interval = 751 ± 2.0595*58/sqrt(26)

Confidence interval = 751 ± 2.0595*11.37473584

Confidence interval = 751 ± 23.4267

Lower limit = 751 - 23.4267 = 727.57

Upper limit = 751 + 23.4267 = 774.43

Confidence interval = (727.57, 774.43)

Effect size is given as below:

d = (Xbar - µ)/S

d = (751 - 747)/58

d = 0.068966

Trivial effect, because difference is less than 0.2.

d) Make an interpretation based on the results.

The amount of sodium in the soup is not significantly different the general listing on the can.

(Because we do not reject the null hypothesis that population mean is 747.)