Answer to Question #333838 in Statistics and Probability for Angel

Question

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?

Answer 1

Let $X=$ the average daily coffee consumption.

The following null and alternative hypotheses need to be tested:

$H_0:\mu\ge3$

$H_1:\mu<3$

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 $\alpha=0.01,$ $df=n-1=20-1=19$ degrees of freedom, and the critical value for a right-tailed test is $t_c = -2.539483.$

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

The t-statistic is computed as follows:

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

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

Using the P-value approach:

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

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

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