Question

In: Math

1. An employment information service claims the mean annual pay for full-time workers without a high...

1. An employment information service claims the mean annual pay for full-time workers without a high school diploma is less than $19,100. The annual pay for a random sample of 8 full-time non-high school graduates is listed below. Salaries for people without high school diplomas are normally distributed. Based on this information, you decide to

18,165             16,012             18,803             19,864

17,328             21,445             19,237             18,316

2. A company that makes cola drinks states that the mean caffeine content per one 12-ounce bottle of cola is 35 milligrams. You believe that the mean caffeine content is not 35 milligrams. You decide to test this claim. During your tests, you find that a random sample of thirty 12-ounce bottles of cola has a mean caffeine content of 38.2 milligrams with a standard deviation of 7.5 milligrams.  Based on this information, you decide to

Solutions

Expert Solution

1)

Below are the null and alternative Hypothesis,
Null Hypothesis: μ = 19100
Alternative Hypothesis: μ < 19100

Rejection Region
This is left tailed test, for α = 0.05 and df = 7
Critical value of t is -1.895.
Hence reject H0 if t < -1.895

Test statistic,
t = (xbar - mu)/(s/sqrt(n))
t = (18646.25 - 19100)/(1634.765/sqrt(8))
t = -0.785

P-value Approach
P-value = 0.2291
As P-value >= 0.05, fail to reject null hypothesis.


2)

Below are the null and alternative Hypothesis,
Null Hypothesis: μ = 35
Alternative Hypothesis: μ ≠ 35

Rejection Region
This is two tailed test, for α = 0.05 and df = 29
Critical value of t are -2.045 and 2.045.
Hence reject H0 if t < -2.045 or t > 2.045

Test statistic,
t = (xbar - mu)/(s/sqrt(n))
t = (38.2 - 35)/(7.5/sqrt(30))
t = 2.337

P-value Approach
P-value = 0.0266
As P-value < 0.05, reject the null hypothesis.


## here, significance level is not given in both i consider 0.05


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