Question

The sodium content of a popular sports drink is listed as 195 mg in a 32-oz bottle. Analysis of 19 bottles indicates a sample mean of 206.4 mg with a sample standard deviation of 22.7 mg.
  
(a) State the hypotheses for a two-tailed test of the claimed sodium content.
  
a. H₀: μ ≥ 195 vs. H₁: μ < 195
b. H₀: μ ≤ 195 vs. H₁: μ > 195

c. H₀: μ = 195 vs. H₁: μ ≠ 195
  

• 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 2 percent level of significance (α = .02), does the sample contradict the manufacturer’s claim?
  
(Click to select)  Do not reject  Reject  H₀. The sample  (Click to select)  contradicts  does not contradict  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)  greater  lower  than the significance level of .02
  
(d-2) Did you come to the same conclusion as you did in part (c)?
  

• Yes

• No

Answer 1

(a) State the hypotheses for a two-tailed test of the claimed sodium content.

c. H0: μ = 195 vs. H1: μ ≠ 195

(b) Calculate the t test statistic to test the manufacturer’s claim. (Round your answer to 4 decimal places.)
  
Test statistic = 2.1891

(c) At the 2 percent level of significance (α = .02), does the sample contradict the manufacturer’s claim?
  
Do not reject H0. The sample does not contradict 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 0.0420. It is greater than the significance level of .02
  
(d-2) Did you come to the same conclusion as you did in part (c)?
  

• Yes

195.000	hypothesized value
206.400	mean 1
22.700	std. dev.
5.208	std. error
19	n
18	df
2.1891	t
.0420	p-value (two-tailed)

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