# Assuming the amount of caffeine consumed per person is normally distributed, is there suffiecient evidence to conclude that the mean amount of caffeine consumed daily by college students is less...

Assuming the amount of caffeine consumed per person is normally distributed, is there suffiecient evidence to conclude that the mean amount of caffeine consumed daily by college students is less than 250mg using a=0.05?

A random sample of 24 college students revealed they consumed on average 226mg of caffeine per day, with a standard deviation of 48mg. Assuming the amount of caffeine consumed per person is normally distributed, is there suffiecient evidence to conclude that the mean amount of caffeine consumed daily by college students is less than 250mg using a=0.05? A. Calculate your test statistic and write down the formula you are using? B. State your decision and explain how reached it? C. State your conclusion?

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We are given a sample with `n=24,bar(x)=226,s=48` . We want to know if `mu<250` with a confidence level of 95%.

Since we are not given the population standard deviation, we use the Student's t-test:

1. `H_0:mu=250` and `H_1:mu<250`

2. To find the critical region we use the Student's t-distribution. The degrees of freedom are 23 (`n-1)` . From the t-distribution table we find the critical value to be -1.71 (`alpha=.05,d.f.=23==>t_(alpha)=-1.71)`

3. The test statistic is found using `t=(bar(x)-mu)/(s/sqrt(n))` . Here we have:

`t=(226-250)/(48/sqrt(24))~~-2.45`

4. Since `-2.45<-1.71` we reject the null-hypothesis. The test value is in the critical region. (Using a calculator I got `p~~.011<.05` so we reject the null hypothesis)

5. There is sufficient evidence to claim that the amoount of caffeine consumed is less than 250mg per day.

*** If the standard deviation given was from the population, then use a z-test. You get the critical value from the z-distribution table and the test value by `z=(bar(x)-mu)/(sigma/sqrt(n))` . The difference will be the critical value. ***

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