Question

The sodium content of a popular sports drink is listed as 250 mg in a 32-oz bottle. Analysis of 16 bottles indicates a sample mean of 260.9 mg with a sample standard deviation of 15.8 mg.

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

a.	H0: μ ≥ 250 vs. H1: μ < 250
b.	H0: μ ≤ 250 vs. H1: μ > 250
c.	H0: μ = 250 vs. H1: μ ≠ 250

	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 (α = .05), does the sample contradict the manufacturer’s claim?

(Reject orDo not reject) H₀. The sample (does not contradict or contradicts) 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 (lower or greater) 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)

H0: = 250

Ha: 250

b)

Test statistics

t = - / S / sqrt(n)

= 260.9 - 250 / 15.8 / sqrt(16)

= 2.7595

This is test statistics value .

c)

Critical value at level for df is 2.131.

Since test statistics value > 2.131, we have sufficient evidence to reject H0.

Reject H0. Sample contradict manufacturer's claim.

d-1)

p value is 0.0146.

(P value formula in excel, = tdist(t,deg_freedom,tail)

= tdist(2.7595,15,2)

= 0.0146)

Since p value < 0.05 significance level, we have sufficient evidence to reject H0.

d-2)

Conclusion for c and d-1 are same.

answered by: anonymous