Question #333838

1. A study shows that the average daily coffee consumption of a 20-30 years old students is 3 cups per day. A university claims that their students tend to drink less than 3 cups. They selected 20 students and found the mean of 3.5 with a standard deviation of 1.5 cups. Use 0.01 level of significance to test their claim.



Answer the following:



Parameter:


Claim:


Claim ( in symbol ):


Ho: Ho:


Ha: Ha:


What is the significance level or a?


Is it two-tailed or one-tailed test?

1
Expert's answer
2022-04-27T09:34:08-0400

Let X=X= the average daily coffee consumption.

The following null and alternative hypotheses need to be tested:

H0:μ3H_0:\mu\ge3

H1:μ<3H_1:\mu<3

This corresponds to a left-tailed test, for which a t-test for one mean, with unknown population standard deviation, using the sample standard deviation, will be used.

Based on the information provided, the significance level is α=0.01,\alpha=0.01, df=n1=201=19df=n-1=20-1=19 degrees of freedom, and the critical value for a right-tailed test is tc=2.539483.t_c = -2.539483.

The rejection region for this left-tailed test is R={t:t<2.539483}.R = \{t: t<-2.539483\}.

The t-statistic is computed as follows:


t=xˉμs/n=3.531.5/201.490712t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{3.5-3}{1.5/\sqrt{20}}\approx1.490712

Since it is observed that t=1.490712>2.539483=tc,t =1.490712 >-2.539483=t_c, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value for left-tailed, df=19df=19 degrees of freedom, t=1.490712t=1.490712 is p=0.923774,p=0.923774, and since p=0.923774>0.01=α,p=0.923774>0.01=\alpha, it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean μ\mu

is less than 3, at the α=0.01\alpha = 0.01 significance level.


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