Answer to Question #204194 in Statistics and Probability for lebo maloke

Question

. A nutritionist is promoting a diet plan for shedding up to 5kg weights within a month without any involvement in comprehensive daily exercises. Some overweight persons had lost 4.1, 2.7, 6.6, 9, 3.8, 6.0, 4.2, 2.0, 3.0, 3.2, 6.3, 2.4, 4.4, 6.2 and 4.5 kgs in a month after following the diet plan. At 5% level of significance, determine whether the efficacy of the diet plan is currently being overrated

Answer 1

$$\bar{x}=\dfrac{1}{15}(4.1+2.7+6.6+9.0+3.8+6.0+4.2$$

$$+2.0+3.0+3.2+6.3+2.4+4.4+6.2+4.5)$$

$$=\dfrac{68.4}{15}=4.56$$

$$s^2=\dfrac{1}{N-1}\sum_i(x_i-\bar{x})^2$$

$$s^2=\dfrac{1}{15-1}\big((4.1-4.56)^2+(2.7-4.56)^2+(6.6-4.56)^2$$

$$+(9.0-4.56)^2+(3.8-4.56)^2+(6.0-4.56)^2$$

$$+(4.2-4.56)^2+(2.0-4.56)^2+(3.0-4.56)^2$$

$$+(3.2-4.56)^2+(6.3-4.56)^2+(2.4-4.56)^2$$

$$+(4.4-4.56)^2+(6.2-4.56)^2+(4.5-4.56)^2)$$

$$=\dfrac{51.576}{14}=3.684$$

$$s=\sqrt{s^2}=\sqrt{3.684}\approx1.919375$$

Hypothesized Population Mean $\mu=5$

Sample Standard Deviation $s=1.919375$

Sample Size $n=15$

Sample Mean $\bar{x}=4.56$

Significance Level $\alpha=0.05$

Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

$H_0: \mu\geq5$

$H_1: \mu<5$

This corresponds to left-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 $\alpha=0.05,$

$df=n-1=14$ ​​degrees of fredom, and the critical value for left-tailed test i$t_c=-1.76131.$

The rejection region for this left-tailed test is $R=\{t:t<-1.76131\}.$

The $t$ - statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{4.56-5}{1.919375/\sqrt{15}}\approx-0.89063$$

Since it is observed that $t=-0.89063>-1.76131=t_c,$ 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 less than $5,$ at the $\alpha=0.05$ significance level.

Using the P-value approach: The p-value for left-tailed, the significance level $\alpha=0.05, df=14, t=-0.89063,$ is $p=0.1940,$ and since $p=0.1940>0.05=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu$ is less than $5,$ at the $\alpha=0.05$ significance level.