Question

I work in an office building and so I asked the question of "how many cups of coffee do you have a week?" I believed that the population mean number of times that my co workers grab coffee each week is less than 3.5. I asked my team of 9 people and found that they grabbed coffee 5,3,4,0,2,1,1,2, and 3 times last week. I choose to test this hypothesis at the .05 significance level.

Does anyone know how to solve this using StatCrunch? Or can help me understand how to test the hypothesis?

Answer 1

Solution:

The first step is to set up the null and alternative hypotheses. The null and alternative hypotheses are:

$H_{0}:\mu\geq 3.5$

$H_{0}:\mu< 3.5$

Under the null hypothesis, the test statistic is:

$t=\frac{\bar{x}-\mu}{\frac{s}{\sqrt{n}}}$

Where:

$\bar{x}=2.33$ is the sample mean given the sample of 9 observations.

$s=1.5811$ is the sample standard deviation of the given sample of 9 observations.

$\therefore t=\frac{2.33-3.5}{\frac{1.5811}{\sqrt{9}}}$

  $=-2.21$

Now we have to find the left tailed p-value because the alternative hypothesis is left tailed.

We can use excel to find the p-value. The excel function is:

$p-value=P(t<-2.21)=TDIST(2.21,8,1)=0.0290$

2.21 is the absolute value of the test statistic, 8 is the degrees of freedom and 1 stand for one-tailed test

Conclusion: Since the p-value is less than the significance level, we, therefore, reject the null hypothesis and conclude that there is sufficient evidence to support the claim that the population mean number of times that my co-workers grab coffee each week is less than 3.5.