In Exercises 14.70-14.80, use the technology of your choice to do the following tasks.
a. Decide whether you can reasonably apply the regression t-test. If so, then also do part (b).
b. Decide, at the 55 significance level, whether the data provide sufficient evidence to conclude that the predictor variable is useful for predicting the response variable.

14.78 Estriol Level and Birth Weight. Use the data on the WeissStats site for estriol levels of pregnant women and birth weights of their children referred to in Exercise 14.42.

To decide whether can reasonably apply the regression t-test. If so, then determine part (b).

The data on the WeissStats site for estriol levels of pregnant women and birth weights of their children referred to in Exercise 14.42 as follows:

The MINITAB is used to create a normal probability plot of residuals.
PROCEDURE FOR MINITAB:
Step 1: Select Stat > Regression > Regression from the drop-down menu.
Step 2: In Response, enter the column Weight
Step 3: In Predictors, enter the columns Estriol.
Step 4: Select columns Estriol under residuals from the Graphs menu.
Step 5: Click the OK button.
The MINITAB output will be:

MINITAB is used to create a normal probability plot of residuals.
PROCEDURE FOR MINITAB:
Step 1: Select Stat > Regression > Regression.
Step 2: In Response, enter the column Weight
Step 3: In Predictors, enter the columns Estriol.
Step 4: Select Normal probability plot of residuals from the Graphs menu.
Step 5: Click the OK button.

The following is the assumption for regression inferences:
Regression line for the population:
For any value of the predictor variable $X$, the conditional mean of the response variable $Y$is ${\beta }_0+{\beta }_{1}X$.
Equal standard deviation:
The response variable's $Y$standard deviation is the same as the explanatory variable's $X$ standard deviation.
The standard deviation is represented by the symbol $\sigma$.
Normal populations:
The response variable's distribution is normally distributed.
Independent observations:
The response variable observations are independent to one another.
Examine the graph for any indications of a violation of one or more of the regression inference assumptions.

• There is an increasing level of variability in the residual plot versus estriol level.
• The residual plot and the normal probability plot of residuals show that the residuals follow a nearly linear pattern.

The regression inferences' normality assumption is not violated here.

As a result, the regression inferences assumption $1-3$ is plausible for the variables weight and Estriol level.

As a result, the regression $t$-test is a reasonable choice for the provided data.

To decide, at the $55$ significance level, whether the data provide sufficient evidence to conclude that the predictor variable is useful for predicting the response variable.

The null hypothesis is indicated as follows:

$H_0:{\beta }_1=0$
To put it another way, the predictor variable "Estriol level" is useless for predicting "weight.
The alternative hypothesis is indicated as follows:
$H_{\alpha }:{\beta }_1\ne 0$
In other words, the predictor variable "Estriol level" can be used to predict "weight."
Rejection Rule:
If $p$is less than $a(0.05)$, reject the null hypothesis $\left(H_0\right)$.
MINITAB can be used to find the test statistic and $p$-value.
PROCEDURE FOR MINITAB:
Step 1: Select Stat > Regression > Regression.
Step 2: In Response, enter the column Weight.
Step 3: In Predictors, enter the columns Estriol.
Step 4: Click the OK button.

The MINITAB output will be:
Regression Analysis: WEIGHT versus ESTRIOL

The test statistic value is $4.14$, and the p-value is $0.000$, according to the MINITAB report.
Use the $\alpha =0.05$significance level. The $p$-value is lower than the level of significance in this case. In other words, the $p$-value is $(=0.000)<\alpha (=0.05)$.
As a result of the rejection rule, it may be argued that at $alpha=0.05$, there is evidence to reject the null hypothesis $\left(H_0\right)$.
Hence, the data support the conclusion that the predictor variable "Estriol level" is beneficial for predicting "weight" at the $5\%$ level.