Answer to Question #337481 in Statistics and Probability for diane

Question

The average amount of rainfall during the summer months is 11.52 inches. A researcher in

PAGASA selects a random sample of 10 provinces and finds that the average amount of

rainfalllast year was 7.42 inches with a standard deviation of 1.3 inches. At 0.01 level

significance, can it be concluded that the mean rainfall last year was below 11.52 inches?

Step

1. State the null and alterative hypothesis

concerning the population mean, $\mu$ and the type of test to be used.

2. Specify the level of significance $\alpha$

3.State the decision rule.

4. Collect the data and perform calculations.

5.Make a statistical decision.

6. State the conclusion.

Answer 1

1. The following null and alternative hypotheses need to be tested:

$H_0:\mu\ge11.52$

$H_a:\mu<11.52$

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.

2. Based on the information provided, the significance level is $\alpha = 0.01,$ $df=n-1=9$ degrees of freedom, and the critical value for a left-tailed test is $t_c = -2.821433.$

3. The rejection region for this left-tailed test is $R = \{t: t <-2.821433\}$

4.The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{7.42-11.52}{1.3/\sqrt{10}}=-9.973337$$

5. Since it is observed that $t = -9.973337<-2.821433= t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for left-tailed, $df=9$ degrees of freedom, $t=-9.973337$ is $p = 0.000002,$ and since $p= 0.000002<0.01=\alpha,$ it is concluded that the null hypothesis is rejected.

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

is less than 11.52, at the $\alpha = 0.01$ significance level.