Answer to Question #342556 in Statistics and Probability for nana

Question

For this test, do the following.

(a) Identify the claim and state H0 and Ha.

(b) Determine whether the hypothesis test is left-tailed, right-tailed, or two-tailed, and whether to use a z-test, a t-test, or a chi-square test. Explain your reasoning.

(c) Choose one of the options. If convenient, use technology.

Option 1: Find the critical value(s), identify the rejection region(s), and find the appropriate standardized test statistic.

Option 2: Find the appropriate standardized test statistic and the P-value.

(d) Decide whether to reject or fail to reject the null hypothesis.

(e) Interpret the decision in the context of the original claim.

2.The U.S. Department of Agriculture claims that the mean annual consumption of tea by a person in the United States is 8.9 gallons. A random sample of 60 people in the United States has a mean annual tea consumption of 8.2 gallons. Assume the population standard deviation is 2.2 gallons. At a = 0.10, can you reject the claim?

Answer 1

a) The following null and alternative hypotheses need to be tested:

$H_0:\mu=8.9$

$H_1:\mu\not=8.9$

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

(c) Based on the information provided, the significance level is $\alpha = 0.1,$ $df=n-1=59$ degrees of freedom, and the critical value for a lefttwo-tailed test is $t_c = 1.671093.$

The rejection region for this two-tailed test is $R = \{t:|t|> 1.671093\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{8.2-8.9}{2.2/\sqrt{2960}}\approx-2.4646$$

Since it is observed that $|t|=2.4646> 1.671093=t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for two-tailed, $df=59$ degrees of freedom, $t=-2.4646,$ is $p= 0.016648,$ and since $p=0.016648<0.1=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$

is different than 8.9, at the $\alpha = 0.1$ significance level.