Answer to Question #204036 in Statistics and Probability for luna granger

Question

 The owner of a mobile phone store in Greenhills shopping center was not happy when he learned that the average number of cellphones sold daily was 8 items only. He hired a new seller and was told to tally the daily sales for 20 days. The new sale records are 9, 9, 9, 8, 9, 9, 11, 12, 12, 16, 10, 12, 8, 11, 17, 18, 15, 14, 17, and 14. by using the default significant level do the collected data enough to indicate that their sale got increased?

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.