Question

The sodium content of a popular sports drink is listed as 205 mg in a 32-oz bottle. Analysis of 20 bottles indicates a sample mean of 219.2 mg with a sample standard deviation of 18.0 mg. (a) State the hypotheses for a two-tailed test of the claimed sodium content. H0: μ ≥ 205 vs. H1: μ < 205 H0: μ ≤ 205 vs. H1: μ > 205 H0: μ = 205 vs. H1: μ ≠ 205

Answer 1

H0: mu equals 205 vs. H1: mu not equals 205.

Explanation:

A null hypothesis (H0) is a statement from a population parameter which is either rejected or accepted (fail to reject) upon testing. It expresses equality.

An alternate hypothesis (H1) is also a statement from the population parameter which negates the null hypothesis and is accepted if the null hypothesis is true. It expresses inequality.

A two-tailed test is one in which the alternate hypothesis is expressed using any of the inequality signs below:

not equal to, less than or equal to, greater than or equal to.

Answer 2

`t=(219.2-205)/((18)/(√(20)))=3.528`

`p_v =2*P(t_(19)>3.528)=0.0022`

If we compare the p value and a significance level for example
`\alpha=0.05` we see that
`p_v <\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the lifetime is signficantly different from 205 hours.

Explanation:

Data given and notation

`\bar X=219.2` represent the sample mean

`s=18` represent the sample standard deviation

`n=20` sample size

`\mu_o =205` represent the value that we want to test

`\alpha` represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

`p_v` represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to apply a two tailed test.

What are H0 and Ha for this study?

Null hypothesis:
`\mu = 205`

Alternative hypothesis :
`\mu \\eq 205`

Compute the test statistic

The statistic for this case is given by:

`t=(\bar X-\mu_o)/((s)/(√(n)))` (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

`t=(219.2-205)/((18)/(√(20)))=3.528`

Give the appropriate conclusion for the test

The degreed of freedom are:

`df = n-1= 19`

Since is a two sided test the p value would be:

`p_v =2*P(t_(19)>3.528)=0.0022`

Conclusion

If we compare the p value and a significance level for example
`\alpha=0.05` we see that
`p_v <\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the lifetime is signficantly different from 205 hours.