Question

The sodium content of a popular sports drink is listed as 220 mg in a 32-oz bottle. Analysis of 10 bottles indicates a sample mean of 228.2 mg with a sample standard deviation of 18.2 mg. (a) State the hypotheses for a two-tailed test of the claimed sodium content. a. H0: μ ≥ 220 vs. H1: μ < 220 b. H0: μ ≤ 220 vs. H1: μ > 220 c. H0: μ = 220 vs. H1: μ ≠ 220 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? H0. The sample 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 than the significance level of 0.05. (d-2) Did you come to the same conclusion as you did in part (c)? Yes No

Answer 1

Answer:

Follows are the responses to this question:

Step-by-step explanation:

$H_O: \mu=220\\\\H_a:\mu \neq 220$

0.05 degree of significance for two tail tests and n-1= 9 df, critica t= 2.262

When the absolute value of the test statistic |t| >2.262, the decision rule rejects$H_0$.

$\mu=220\\\bar{x}=228.200\\n=10.00\\smaple \ std \ deviation = 18.200\\std\ error \ ' sx= \frac{s}{\sqrt{n}} =5.7553\\test stat t=\frac{(x-\mu) \times \sqrt{n}}{sx}=1.4248\\p value=0.1880$

In point c:

It does not reject$H_0$, because it has the sample that not contradicts the manufacturer's claim

In point d:

For point 1:  

The value of p is 0.1880 , that is higher than the 0.05

For point 2:

Yes