Question

The sodium content of a popular sports drink is listed as 202 mg in a 32-oz bottle. Analysis of 17 bottles indicates a sample mean of 213.7 mg with a sample standard deviation of 16.4 mg.
  
(a) State the hypotheses for a two-tailed test of the claimed sodium content.
  
a. H₀: ? ? 202 vs. H₁: ? < 202
b. H₀: ? ? 202 vs. H₁: ? > 202

c. H₀: ? = 202 vs. H₁: ? ? 202
  

	a
	b
	c

  
(b) Calculate the t test statistic to test the manufacturer’s claim. (Round your answer to 4 decimal places.)
  
Test statistic            
  
(c) At the 5 percent level of significance (? = .05), does the sample contradict the manufacturer’s claim?
  
(Click to select)RejectDo not reject H₀. The sample (Click to select)does not contradictcontradicts the manufacturer’s claim.
  
(d-1) Use Excel to find the p-value and compare it to the level of significance. (Round your answer to 4 decimal places.)
  
The p-value is . It is (Click to select)lowergreater than the significance level of .05
  
(d-2) Did you come to the same conclusion as you did in part (c)?
  

	Yes
	No

Answer 1

a) The first two options are lower tailed and upper tailed test both being one tailed tests. Therefore c) is the correct answer here.

c. H0: ? = 202 vs. H1: ? ? 202

b) The test statistic here is computed as:

Therefore 2.9415 is the test statistic value here.

c) The p-value here is computed for n - 1 = 16 degrees of freedom, from the t distribution tables as

p = 2P( t₁₆ > 2.9415 ) = 2*0.0048 = 0.0096

Now as the p-value here is 0.0096 < 0.05 which is the level of significance, therefore the test is significant and we can reject the null hypothesis here. Reject H_o. Yes it contradicts the claim.

(d-1) This was already computed in previous part as being equal to 0.0096

(d -2) Yes we come to the same conclusion here.