Answer to Question #260276 in Statistics and Probability for yyy

Question

According to an officer in the Health Care Financing Administration, the average

annual expenditure on toiletries is RM110 per person in Malaysia. The data of 30

randomly selected persons are as follows and the standard deviation of the sample is

19.2152.

125 110 110 90 90

85 125 110 85 120

130 125 130 90 130

80 120 90 110 90

90 85 100 130 80

90 100 125 150 100

i. Find the sample mean of the data.

ii. Is there enough evidence to support the officer’s claim at 0.02 level of significance?

iii. What is the decision at a 0.2% significance level?

Answer 1

i.

$$mean=\bar{x}=\dfrac{1}{n}\sum_i=\dfrac{1}{30}(125+110 +110+ 90$$

$$+90+85+ 125+ 110+ 85 +120+130+ 125$$

$$+130+ 90+ 130+80+ 120+ 90 +110 +90$$

$$+90+ 85+ 100+ 130+ 80+90+ 100+ 125$$

$$+ 150+ 100)=106.5$$

ii.

The following null and alternative hypotheses need to be tested:

$H_0: \mu=110$

$H_1: \mu\not=110$

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 $\alpha = 0.02,df=n-1=29$ degrees of freedom, and the critical value for a left-tailed test is $t_c = 2.462021.$

The rejection region for this left-tailed test is $R = \{t: |t| > 2.462021\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{106.5-110}{19.2152/\sqrt{30}}=-0.99766$$

Since it is observed that $|t| = 0.99766 \le2.462021= t_c ,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value for two-tailed, $\alpha=0.02, df=29, t=-0.99766$ is $p=0.326696,$ and since $p=0.326696>0.02=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu$ is different than 110, at the  $\alpha = 0.02$ significance level.

iii.

Using the P-value approach: The p-value for two-tailed, $\alpha=0.002, df=29, t=-0.99766$ is $p=0.326696,$ and since $p=0.326696>0.002=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu$ is different than 110, at the  $\alpha = 0.002$ significance level.