Question

Use the data on the Weiss Stats site for estriol levels of pregnant women and birth weights of their children referred to in Exercise 14.42.

Short answer

(a) For the provided data, the regression t-test is appropriate.

(b) The data support the conclusion that the predictor variable "Estriol level" is beneficial in predicting "weight" at the $5\%$level.

Step-by-step solution

Part (a) Step 1: Given information

Given in the question that, Use the data on the Weiss Stats site for estriol levels of pregnant women and birth weights of their children referred to in Exercise 14.42. We need to decide that whether we can reasonably apply the regression t-lest. If so, then also do part (b).

Part (a) Step 2: Explanation

Given:

MINITAB is used to create the residual plot.

Procedure for MINITAB: Step 1: Select Stat >Regression > Regression from the drop-down menu.

Step 2: In the Response column, type Weight.

Step 3: In Predictors, fill in the Estriol columns.

Step 4: In Graphs, under Residuals vs the variables, enter the columns Estriol.

Step 5: Click the OK button.

OUTPUT FROM MINITAB:

Part (a) Step 3: Normal probability plot of residuals 

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: Fill in the Weight column in Response.

Step 3: In Predictors, fill in the Estriol columns.

Step 4: Select Normal probability plot of residuals from the Graphs menu.

Step 5: Click the OK button.

OUTPUT FROM MINITAB:

The following is the assumption for regression inferences:

Line of population regression:

For each value $X$of the predicator variable, the conditional mean of the response variable $\left(Y\right)$is $\beta 0+\beta 1X$.

Standard deviations are equal:

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 $\sigma$.

Typical populations include:

The response variable follows a normal distribution.

Observations made independently:

The response variable observations are unrelated to one another.

Examine whether the graph shows 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 trend.

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.

Part (b) Step 1: Given information

Given in the question that, Use the data on the Weiss Stats site for estriol levels of pregnant women and birth weights of their children referred to in Exercise 14.42. we need to decide, at the $5\%$significance level, whether the data provide sufficient evidence to conclude that the predictor variable is useful for predicting the response variable.

Part (b) Step 2: Explanation

The following are the suitable hypotheses:

Hypothesis of nullity:

$H_0:{\beta }_1=0$

In other words, the predictor variable "Estriol level" is ineffective in predicting "weight."

Another possibility:

$H_{\alpha }:{\beta }_1\ne 0$

In other words, the predictor variable "Estriol level" can be used to predict "weight."

If the $p$-value is less than or equal to 0.05, reject the null hypothesis $\left(H_0\right)$

MINITAB can be used to find the test statistic and p-value.

Step 1: Select Stat > Regression > Regression in MINITAB.

Step 2: In the Response column, type Weight.

Step 3: In Predictors, fill in the Estriol columns.

Step 4: Click the OK button.

OUTPUT FROM MINITAB:

WEIGHT vs. ESTRIOL: A Regression Analysis

The test statistic value is 4.14, and the p-value is $0.000$, according to the MINITAB report.

Conclusion: Use the $\alpha =0.05$significance threshold.

The significance level is less than the $p$-value.

In other words, the p-value$(=0.000)<\alpha (=0.05)$

According to the rejection criterion, there is sufficient evidence to reject the null hypothesis ($\left.H_0\right)$at $\alpha =0.05$.