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 ≠ 200H0: µ ≥ 200 vs. H1: µ < 200H0: 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?__(reject or do not reject)____ H0. The sample _____(contradict or does not contradict)______ 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)?

Answer: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 mgSample mean, [tex]\bar{x}[/tex] = 211.5 mgSample size, n = 20Alpha, α = 0.05 Sample standard deviation, s = 18.5 mga) First, we design the null and the alternate hypothesis  for a two tailed test[tex]H_{0}: \mu = 200\\H_A: \mu \neq 200[/tex] We use Two-tailed t test to perform this hypothesis.b) Formula:[tex]t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }[/tex] Putting all the values, we have [tex]t_{stat} = \displaystyle\frac{211.5 - 200}{\frac{18.5}{\sqrt{20}} } = 2.7799[/tex] c) Now, [tex]t_{critical} \text{ at 0.05 level of significance, 19 degree of freedom } = \pm 2.0930[/tex] 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 = 0.011934Since 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.