A researcher conducts an experiment to examine the relationship between the weig

Question

A researcher conducts an experiment to examine the relationship between the weight gain (wtgn, gm) of chickens, whose diets had been supplemented by different amounts of lysine, and the amount of lysine (gm) ingested. A random sample of 12 chicks was selected for the study. The R output is given below Call: 1m (formula-wtgnlysine, data -lys) Coefficients: Estimate Std. Error t value Pr >tl) 1.1 10.65 8.9e-07 6.99 5.15 0.00043 12.57 (Intercept) lysine Residual standard error: 1.02 on 10 degrees of freedom Multiple R-squared: 0.726,Adjusted R-squared: 0.699 F-statistic: 26.5 on 1 and 10 DE, p-value: 0.000432 35.48 95% CI fit lw upr 16.9 15.9 17.8 95% PI fit lwr upr 16.8 14.4 19.3 The results for a test of Ho : 1 = 0 versus Ha : 10 show that: , the null hypothesis can be rejected because t=5.15, p=0.00043 o the null hypothesis cannot be rejected because t=10.65, p=8.9e-07 o the null hypothesis cannot be rejected because t=5.15, p=0.00043 the null hypothesis can be rejected because t 10.65, p=8.9e-07

Explanation / Answer

Solution:-

The null hypothesis can be rejected because t = 5.15, p = 0.00043

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

H0: The slope of the regression line is equal to zero.
Ha: The slope of the regression line is not equal to zero.

Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a linear regression t-test to determine whether the slope of the regression line differs significantly from zero.

Analyze sample data. To apply the linear regression t-test to sample data, we require thestandard error of the slope, the slope of the regression line, the degrees of freedom, the t statistic test statistic, and the P-value of the test statistic.

We get the slope (b1) and the standard error (SE) from the regression output.

b1 = 35.48 SE = 6.89

We compute the degrees of freedom and the t statistic test statistic, using the following equations.

DF = n - 2

D.F = 10

t = b1/SE

t = 5.15

where DF is the degrees of freedom, n is the number of observations in the sample, b1 is the slope of the regression line, and SE is the standard error of the slope.

Based on the t statistic test statistic and the degrees of freedom, we determine the P-value

Therefore, the P-value is 0.000432

Interpret results. Since the P-value (0.000432) is less than the significance level (0.05), we cannot accept the null hypothesis.

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