Question

A coffee machine dispenses coffee into paper cups. You are supposed to get 10 ounces of coffee, but the amount varies slightly from cup to cup. Here are the amounts measured in a random sample of 20 cups:

9.9               9.7             10.0             10.1             9.9   

                         9.6               9.8             9.8             10.0             9.5

                         9.7               10.1             9.9             9.6             10.2                                                9.8               10.0             9.9             9.5        9.9

                         

                         

1. Use StatCrunch to construct a frequency histogram that corresponds to the given data. Can a t-distribution be used? Explain why or why not.

2. Find the mean and the standard deviation of the sample using Minitab.

The sample mean is:

The sample standard deviation is:

3. Using this data, by hand, estimate the true population mean fill of coffee with 95% confidence. You must provide a detailed graph, the margin of error and the confidence interval endpoints for credit. Then confirm with MINITAB. Interpret your results with a meaningful answer.

PART II: Keep your glorious piece of work from the confidence interval for the following hypothesis test:

4.    At the 5% significance level, is there sufficient evidence that the average fill of coffee is less than 10 ounces? In other words: Is the machine shortchanging customers? Explain.

Answer 1

Since data is approximately normal and sample size is less than 30, thus we can use t- distribution.

Using Minitab.

The sample mean is: 9.845

The sample standard deviation is: 0.1986

We need to construct the 95% confidence interval for the population mean \muμ. The following information is provided:

The number of degrees of freedom are df=20−1=19, and the significance level is α=0.05.

Based on the provided information, the critical t-value for α=0.05 and =19 degrees of freedom is t_c​=2.093.

  

  

Now, we want to test if there sufficient evidence that the average fill of coffee is less than 10 ounces:

The following null and alternative hypotheses need to be tested:

H_o: μ = 10

H_a: μ < 10

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

The t-statistic is computed as follows:

  

Based on the information provided, the significance level is α=0.05, and the critical value for a left-tailed test is

t_c​=−1.729.

Since it is observed that t=−3.49 < t_c​=−1.729, it is then concluded that the null hypothesis is rejected.

Hence it is concluded that there are  sufficient evidence that the average fill of coffee is less than 10 ounces.