Question

3. The owner of a cell phone store at Greenhill Shopping Center was not happy when he learned that the

average number of cell phones sold daily was only 8 mobile phones. He hired a new seller and told him to

tally the daily sales for 20 days. The following are the number of cell phones sold for the past 20 days.

9

11

12

6

12

8

17

14

10

12

11

18

15

14

17

Does the collected data present sufficient evidence to indicate that the average number of cell phones sold

daily has increased? Use 0.05 level of significance. Assume normality in the population.

Answer 1

$$9, 9, 9, 8, 9, 9, 11, 12, 12, 16,$$ $$10, 12, 8, 11, 17, 18, 15, 14, 17,14$$ $$n=20, \sum_ix_i=240$$ $$\bar{x}=\dfrac{1}{20}\sum_ix_i=\dfrac{240}{20}=12$$ $$\sum_i(x_i-\bar{x})^2=(9-12)^2+(9-12)^2+(9-12)^2$$ $$+(8-12)^2+(9-12)^2+(9-12)^2+(11-12)^2$$ $$+(12-12)^2+(12-12)^2+(16-12)^2+(10-12)^2$$ $$+(12-12)^2+(8-12)^2+(11-12)^2+(17-12)^2$$ $$+(18-12)^2+(15-12)^2+(14-12)^2+(17-12)^2$$ $$+(14-12)^2=202$$ $$s^2=\dfrac{\sum_i(x_i-\bar{x})^2}{n-1}=\dfrac{202}{20-1}=\dfrac{202}{19}$$ $$\approx10.631579$$ $$s=\sqrt{s^2}\approx3.260610$$

Hypothesized Population Mean $\mu=8$

Sample Standard Deviation $s=3.260610$

Sample Size $n=20$

Sample Mean $\bar{x}=12$

Significance Level $\alpha=0.05$

Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

$H_0: \mu\leq8$

$H_1: \mu>8$

This corresponds to right-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 $\alpha=0.05,$ $df=n-1=19$ degrees of freedom and the critical value for right-tailed test is $t_c=1.729133.$

The rejection region for this right-tailed test is $R=\{t:t>1.729133\}.$

The $t$ - statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{12-8}{3.260610/\sqrt{20}}\approx5.486257$$

Since it is observed that $t=5.486257>1.729133=t_c,$ it is then concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$ is greater than $8,$ at the $\alpha=0.05$ significance level.

Using the P-value approach: The p-value for right-tailed, the significance level 

$\alpha=0.05, t=5.486257, df=19$ is $p=0.000014,$ and since $p=0.000014<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$ is greater than $8,$ at the $\alpha=0.05$ significance level.