Answer to Question #344838 in Statistics and Probability for Husmon

Question

. The hospital record shows that the mean weight of newly born baby is greater than 7 lbs. A researcher takes a sample of 25 newly born babies and found to have a mean weight of 6.97lbs with the standard deviation of 0.77 lbs. Test the claim at 0.05 level of significance.

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu\le 7$

$H_1:\mu>7$

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 $\alpha = 0.05,$ $df=n-1=24$ and the critical value for a right-tailed test is $t_c =1.710882.$

The rejection region for this right-tailed test is $R = \{t:t>1.710882\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{6.97-7}{0.77/\sqrt{25}}=-0.1948$$

Since it is observed that $t=-0.1948<1.710882=t_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value for right-tailed, $df=24$ degrees of freedom, $t=-0.1948$ is $p=0.576405,$ and since $p=0.576405>0.05=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu$

is greater than 7, at the $\alpha = 0.05$ significance level.