Answer to Question #142765 in Statistics and Probability for mbk

Question

A research firm conducted a survey to determine the mean amount of money smokers spend on cigarettes during a day. A sample of 100 smokers revealed that the sample mean is N$ 5.24 and sample standard deviation is N$ 2.18 Assume that the sample was drawn from a normal population. 2.5) Use the data provided above to test a claim at a 1 % Level of Significance that the unknown population mean of spending amount on cigarettes is N$ 5.00 (8)

Answer 1

The provided sample mean is $\bar{x}=5.24$ and the sample standard deviation is $s=2.18,$ and the sample size is $n=100.$

The following null and alternative hypotheses need to be tested:

$H_0:\mu=5$

$H_1:\mu\not=5$

This corresponds to a two-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used.

The number of degrees of freedom are $df=100-1=99.$

Based on the information provided, the significance level is $\alpha=0.01,$ and the critical value for a two-tailed test is $t_c=2.626405$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{5.24-5}{2.18/\sqrt{100}}\approx1.100917$$

Since it is observed that $|t|=1.100917<2.626405=t_c,$ it is then concluded that the null hypothesis is not rejected. Therefore, there is not enough evidence to claim that the population mean $\mu$  is different than 5, at the 0.01 significance level.

Using the P-value approach: The p-value is $p=0.2736,$ and since $p=0.2736>0.01,$ 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 5, at the 0.01 significance level.