The sodium content of a popular sports drink is listed as 200 mg in a 32-oz bottle. analysis of 20 bottles indicates a sample mean of 211.5 mg with a sample standard deviation of 18.5 mg.

(a)
state the hypotheses for a two-tailed test of the claimed sodium content.
h0: µ = 200 vs. h1: µ ≠ 200
h0: µ ≥ 200 vs. h1: µ < 200
h0: µ ≤ 200 vs. h1: µ > 200

(b)
calculate the t test statistic to test the manufacturer’s claim.

(c)
at the 5 percent level of significance (α = 0.05) does the sample contradict the manufacturer’s claim?
or do not h0. the sample or does not the manufacturer's claim.

(d-1)
use excel to find the p-value and compare it to the level of significance.
did you come to the same conclusion as you did in part (c)?

The sample does not contradicts the manufacturer's claim.

Step-by-step explanation:

We are given the following in the question:  

Population mean, μ = 200 mg

Sample mean, = 211.5 mg

Sample size, n = 20

Alpha, α = 0.05

Sample standard deviation, s = 18.5 mg

a) First, we design the null and the alternate hypothesis  for a two tailed test

We use Two-tailed t test to perform this hypothesis.

b) Formula:

Putting all the values, we have

c) Now,

Since, the calculated t-statistic does not lie in the range of the acceptance region(-2.0930,2.0930), we reject the null hypothesis.

Thus, the sample does not contradicts the manufacturer's claim.

d) P-value = 0.011934

Since the p-value is less than the significance level, we reject the null hypothesis and accept the alternate hypothesis.

Yes, both approach the critical value and the p-value approach gave the same results.

13.33%

say 50 oranges, then 100 apples

80 are red, 20 are yellow

total fruit = 150

20chances out of 150 for yellow

20/150 = 13.33%