Question

The quality control department of a soup brand is concerned that one of its soups contains less sodium than the 752 mg per serving listed on the can. A sample of 14 cans from the canning facility had an average sodium of 801 mg and a standard deviation of 65.3 mg per serving. What can be concluded with α = 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

b)
Population:
---Select--- quality control department sodium listed on can cans from facility sodium soup brand
Sample:
---Select--- quality control department sodium listed on can cans from facility sodium soup brand

c) 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

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

e) 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

f) 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

Solution:-

a) One sample t-test.

b)

Population:  All the soups from the canning facility of a soup brand.
Sample: 14 cans from the canning facility had an average sodium of 801 mg and a standard deviation 65.3 mg per serving.

c)

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

Null hypothesis: u > 752
Alternative hypothesis: u < 752

Note that these hypotheses constitute a one-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample t-test.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = s / sqrt(n)

S.E = 17.4522
DF = n - 1

D.F = 14
t = (x - u) / SE

t = 2.81

t_Critical = - 1.761

Rejection region is t < - 1.761

where s is the standard deviation of the sample, x is the sample mean, u is the hypothesized population mean, and n is the sample size.

Interpret results. Since the t-value (2.81) does not lies in the rejection region, hence we failed to reject the null hypothesis.

From the above test we have sufficient evidence in the favor of the claim that one of its soups contains less sodium than the 752 mg per serving listed on the can.

d) 95% confidence interval for the population mean is C.I = ( 714.565, 789.435).

C.I = 752 + 2.145 × 17.4522

C.I = 752 + 37.435

C.I = ( 714.565, 789.435)