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A recent study revealed the typical American coffee drinker consumes an average of 3.1 cups per day. A sample of 12 senior citizens revealed they consumed the following amounts of coffee, reported in cups, yesterday.
3.1, 3.3 ,3.5 ,2.6, 2.6, 4.3, 4.4 ,3.8 ,3.1 ,4.1 ,3.1, 3.2
(a) State the null hypothesis and the alternate hypothesis. (Round your answers to 1 decimal place.)
(b) State the decision rule for 0.05 significance level. (Negative amounts should be indicated by a minus sign. Round your answers to 3 decimal places.)
(c) Compute the value of the test statistic. (Round your answer to 3 decimal places.)
(d) Do these sample data suggest there is a difference between the national average and the sample mean of the senior citizens?

Sagot :

Using the t-distribution, it is found that:

a)

The null hypothesis is [tex]H_0: \mu = 3.1[/tex]

The alternative hypothesis is [tex]H_1: \mu \neq 3.1[/tex]

b)

  • |t| < 2.201: Do not reject the null hypothesis.
  • |t| > 2.201: Reject the null hypothesis.

c) t = 1.853

d) Since |t| = 1.853 < 2.2, we do not reject the null hypothesis, that is, the sample data does not suggest that there is a difference between the national average and the sample mean of the senior citizens.

Item a:

At the null hypothesis, it is tested if the estimate of 3.1 cups per day is correct, that is:

[tex]H_0: \mu = 3.1[/tex]

At the alternative hypothesis, it is tested if the estimate is not correct, that is:

[tex]H_1: \mu \neq 3.1[/tex]

Item b:

This is a two-tailed test, as we are testing if the mean is different of a value, with 12 - 1 = 11 df and a significance level of 0.05,  hence, the critical value is [tex]t^{\ast} = 2.2[/tex].

Then, the decision rule is:

  • |t| < 2.201: Do not reject the null hypothesis.
  • |t| > 2.201: Reject the null hypothesis.

Item c:

We can find the standard deviation for the sample, hence, the t-distribution is used.

The test statistic is given by:

[tex]t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}[/tex]

The parameters are:

  • [tex]\overline{x}[/tex] is the sample mean.
  • [tex]\mu[/tex] is the value tested at the null hypothesis.
  • s is the standard deviation of the sample.
  • n is the sample size.

The values of the parameters, with the help of a calculator for the sample mean and standard deviation, are given by: [tex]\overline{x} = 3.425, \mu = 3.1, s = 0.6077, n = 12[/tex]

Hence, the value of the test statistic is:

[tex]t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}[/tex]

[tex]t = \frac{3.425 - 3.1}{\frac{0.6077}{\sqrt{12}}}[/tex]

[tex]t = 1.853[/tex]

Item d:

Since |t| = 1.853 < 2.2, we do not reject the null hypothesis, that is, the sample data does not suggest that there is a difference between the national average and the sample mean of the senior citizens.

A similar problem is given at https://brainly.com/question/24826023

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