Answer to Question #217735 in Statistics and Probability for Madimetja

Question

According to a dietician, adult South African men are on average more than 10kg overweight. To test this statement, twenty South African men are chosen at random. Their weights (kg) are measured and their respective ideal weights subtracted. The results are:

8 11.5 9 20.5 11 9 11.5 9 11 7.5

9 17.5 8 7.5 8.5 9.5 11.5 7.5 8 13

It follows that x =10:4kg and s = 3:3896kg. Investigate the statement of the dietician

(a) With a 95% confidence interval; write down the formula and calculate the realised interval estimate.

(b) Test the dietician’s claim at the 2:5% level of significance.

Answer 1

(a)

$$mean=\bar{x}=\dfrac{1}{20}(8 +11.5+ 9+ 20.5+ 11 +9$$

$$+11.5+ 9+ 11+ 7.5+9 +17.5+ 8 +7.5 +8.5$$

$$+ 9.5+ 11.5+ 7.5 +8 +13)=10.4$$

$$Variance=s^2=\dfrac{1}{20-1}((8-10.4)^2$$

$$+(11.5-10.4)^2+(9-10.4)^2+(20.5-10.4)^2$$

$$+(11-10.4)^2+(9-10.4)^2+(11.5-10.4)^2$$

$$+(9-10.4)^2+(11-10.4)^2+(7.5-10.4)^2$$

$$+(9-10.4)^2+(17.5-10.4)^2+(8-10.4)^2$$

$$+(7.5-10.4)^2+(8.5-10.4)^2+(9.5-10.4)^2$$

$$+(11.5-10.4)^2+(7.5-10.4)^2+(8-10.4)^2$$

$$+(13-10.4)^2\approx11.489474$$

$$s=\sqrt{s^2}\approx 3.3896$$

The critical value for $\alpha=0.05$ and $df=n-1=20-1=19$ degrees of freedom is $t_c=z_{1-\alpha/2, n-1}= 2.093024.$

The corresponding confidence interval is computed as shown below:

$$CI=(\bar{x}-t_c\times\dfrac{s}{\sqrt{n}}, \bar{x}+t_c\times\dfrac{s}{\sqrt{n}})$$

$$=(10.4-2.0930\times\dfrac{3.3896}{\sqrt{20}}, 10.4+2.0930\times\dfrac{3.3896}{\sqrt{20}})$$

$$=(8.8136, 11.9864)$$

Therefore, based on the data provided, the 95% confidence interval for the population mean is $8.8136<\mu< 11.9864,$ which indicates that we are 95% confident that the true population mean $\mu$ is contained by the interval $(8.8136, 11.9864 ).$

(b) The following null and alternative hypotheses need to be tested:

$H_0:\mu\leq10$

$H_1:\mu>10$

This corresponds to a 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 $\alpha=0.025, df=19$ degrees of freedom, and the critical value for a right-tailed test is $t_c=2.093024.$

The rejection region for this right-tailed test is $R=\{t:t>2.093024\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{10.4-10}{3.3896/\sqrt{20}}=0.527748$$

Since it is observed that$t=0.527748<2.093024=t_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value for right-tailed, $\alpha=0.025, df=19,$

$t=0.527748$ is $p=0.301892,$ and since $p=0.301892>0.025=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu$ is greater than $10,$ at the $\alpha=0.025$ significance level.