Answer to Question #260278 in Statistics and Probability for yyy

Question

The birth weight of babies is normally distributed with a mean of 3.5 kg. A researcher

suspects that the weight of babies from mothers who were alcoholics during their

pregnancy will have a lower weight compared to the population mean. A random

sample of 40 babies from such mothers who drank a lot of alcohol were examined. The

mean birth weight in this sample was 3.2 kg with a standard deviation of 0.5 kg.

i. Would you conclude that the researcher’s claim is true at a 1% significance level?

ii. Make a 94% confidence interval for the mean birth weight of all babies.

Answer 1

i. The following null and alternative hypotheses need to be tested:

$H_0: \mu\geq 3.5$

$H_1: \mu<3.5$

This corresponds to a 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 $\alpha = 0.01,df=n-1=39$ degrees of freedom, and the critical value for a left-tailed test is $t_c =-2.425841.$

The rejection region for this left-tailed test is $R = \{t: t <-2.425841\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{3.2-3.5}{0.5/\sqrt{40}}=-3.7947$$

Since it is observed that $t = -3.7947 \le-2.425841= t_c ,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value for left-tailed, $\alpha=0.01, df=39, t=-3.7947$ is $p=0.000252,$ and since $p=0.000252<0.01=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$ is less than 3.5, at the  $\alpha = 0.01$ significance level.

Therefore, there is enough evidence to claim that the weight of babies from mothers who were alcoholics during their pregnancy will have a lower weight, at the  $\alpha = 0.025$ significance level.

ii.The critical value for $\alpha = 0.06$ and $df = n-1 = 39$ degrees of freedom is$t_c = z_{1-\alpha/2; n-1} =1.937106.$

The corresponding confidence interval is computed as shown below:

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

$$=(3.2-1.937106\times\dfrac{0.5}{\sqrt{40}}, 3.2+1.937106\times\dfrac{0.5}{\sqrt{40}})$$

$$=(3.047, 3.353)$$

Therefore, based on the data provided, the 94% confidence interval for the population mean is $3.047 < \mu < 3.353,$ which indicates that we are 92% confident that the true population mean $\mu$

is contained by the interval $(3.047,3.353).$