Answer to Question #348522 in Statistics and Probability for adrian

Question

The claim is made that Internet shoppers spend on the average $335 per year. It is desired to test that this figure is not correct at a = 0.075. Three hundred Internet shoppers are surveyed and it is found that the sample mean = $354 and the standard deviation = $125. Find the value of the test statistic and give your conclusion. Note: critical value is +1.7805 and -1.7805.

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=335$

$H_1:\mu\not=335$

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.

Based on the information provided, the significance level is $\alpha = 0.075,$ $df=n-1=299$ and the critical value for a two-tailed test is $t_c =1.786694.$

The rejection region for this two-tailed test is $R = \{t:|t|>1.786694\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{354-335}{125/\sqrt{300}}=2.6327$$

Since it is observed that $|t|=2.6327>1.786694=t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for two-tailed, $df=299$ degrees of freedom, $t=2.6327$ is $p=0.008911,$ and since $p= 0.008911<0.075=\alpha,$ it is concluded that the null hypothesis is rejected.

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

is different than 335, at the $\alpha = 0.075$ significance level.