Chapter 13, Problem 31CE

To determine

Find the degrees of freedom.

Obtain the critical value.

State the conclusion.

Answer to Problem 31CE

The degrees of freedom are 68.

The critical value is $\pm 1.995$.

The predictor ‘BachDeg%’ is significant.

Explanation of Solution

Calculation:

The given information is that, the dataset of ‘Noodles & Company Sales, Seating, and Demographic data’ contains $n=74$ observations. The response variable is ‘annual sales per square foot’, there are $k=5$ predictor variables ‘Interior Seat Count, Patio Seat Count, Median HH Income, Median Age of Population, % with Bachelor's Degree’.

The formula for the test statistic for coefficient of predictor $X_j$ is,

$t_{calc}=\frac{b_j-0}{s_j}$

Where, $b_j$ is the coefficient, and $s_j$ is the standard error.

The formula for degrees of freedom is,

$df=n-k-1$

Where, n is the total number of observations, k is the number of predictors.

Software procedure:

Step by step procedure to obtain test statistics using MegaStat software is given as,

• • Choose MegaStat >Correlation/Regression>Regression Analysis.
• • SelectInput ranges, enter the variable range for ‘Seats-Inside, Seats-Patio, MedIncome, MedAge, BachDeg%’ as the column of X, Independent variable(s)
• • Enter the variable range for ‘Sales/SqFt’ as the column of Y, Dependent variable.
• • Click OK.

Output using MegaStatsoftware is given below:

The test statistic for predictor seats-inside is –1.819.

The test statistic for predictor seats-patio is 1.198.

The test statistic for predictor MedIncome is –1.921.

The test statistic for predictor MedAge is –0.004.

The test statistic for predictor BachDeg% is 3.307.

Substitute 74 for n, and 5 for k in the degrees of freedom formula.

$\begin{array}{c}df=74-5-1\\ =74-6\\ =68\end{array}$

The degrees of freedom are 68.

Software procedure:

Step by step procedure to obtain the critical value using MINITAB software is given as,

• • Choose Graph > Probability Distribution Plot choose View Probability > OK.
• • From Distribution, choose ‘t’ distribution.
• • In Degrees of freedom, enter 68.
• • Click the Shaded Area tab.
• • Choose probability and Two Tail for the region of the curve to shade.
• • Enter the data value as 0.05.
• • Click OK.

The output of the software is,

Hence, the critical value for 68 degrees of freedom with 0.05, level of significance is $\pm 1.995$.

Decision rules:

• • If test statistic value lies between positive and negative critical value, then do not reject null hypothesis.
• • Otherwise, reject the null hypothesis.

Coefficient of Seats-Inside:

$H_0:{\beta }_1=0$

That is variable Seats-Inside is not related to Sales/SqFt.

Alternative hypothesis:

$H_1:{\beta }_1\ne 0$

That is variable Seats-Inside is related to Sales/SqFt.

Conclusion:

The value of test statistic is –1.819.

The critical values are –1.995 and 1.995.

The test statistic value lies between positive and negative critical value.

That is, $-1.995\left(=-t_{\frac{\alpha }{2},n-1}\right)<-1.819\left(=t_{\text{cal}}\right)<1.995\left(=t_{\frac{\alpha }{2},n-1}\right)$

Hence the null hypothesis is not rejected.

The predictor Seats-Inside is not significant.

Coefficient of Seats-Patio:

$H_0:{\beta }_2=0$

That is variable Seats-Patio is not related to Sales/SqFt.

Alternative hypothesis:

$H_1:{\beta }_2\ne 0$

That is variable Seats-Patio is related to Sales/SqFt.

Conclusion:

The value of test statistic is 1.198.

The critical values are –1.995 and 1.995.

The test statistic value lies between positive and negative critical value.

That is, $-1.995\left(=-t_{\frac{\alpha }{2},n-1}\right)<1.198\left(=t_{\text{cal}}\right)<1.995\left(=t_{\frac{\alpha }{2},n-1}\right)$

Hence the null hypothesis is not rejected.

The predictor Seats-Patio is not significant.

Coefficient of MedIncome:

$H_0:{\beta }_3=0$

That is variable MedIncome is not related to Sales/SqFt.

Alternative hypothesis:

$H_1:{\beta }_3\ne 0$

That is variable MedIncome is related to Sales/SqFt.

Conclusion:

The value of test statistic is –1.921.

The critical values are –1.995 and 1.995.

The test statistic value lies between positive and negative critical value.

That is, $-1.995\left(=-t_{\frac{\alpha }{2},n-1}\right)<-1.921\left(=t_{\text{cal}}\right)<1.995\left(=t_{\frac{\alpha }{2},n-1}\right)$

Hence the null hypothesis is not rejected.

The predictor MedIncome is not significant.

Coefficient of MedAge:

$H_0:{\beta }_4=0$

That is variable MedAge is not related to Sales/SqFt.

Alternative hypothesis:

$H_1:{\beta }_4\ne 0$

That is variable MedAge is related to Sales/SqFt.

Conclusion:

The value of test statistic is –0.004.

The critical values are –1.995 and 1.995.

The test statistic value lies between positive and negative critical value.

That is, $-1.995\left(=-t_{\frac{\alpha }{2},n-1}\right)<-0.004\left(=t_{\text{cal}}\right)<1.995\left(=t_{\frac{\alpha }{2},n-1}\right)$

Hence the null hypothesis is not rejected.

The predictor MedAge is not significant.

Coefficient of BachDeg%:

$H_0:{\beta }_5=0$

That is variable BachDeg% is not related to Sales/SqFt.

Alternative hypothesis:

$H_1:{\beta }_5\ne 0$

That is variable BachDeg% is related to Sales/SqFt.

Conclusion:

The value of test statistic is 3.307.

The critical values are –1.995 and 1.995.

The test statistic value is greater than the positive critical value.

The test statistic value does not lie between positive and negative critical value.

That is, $3.307\left(=t_{\text{cal}}\right)>1.995\left(=t_{\frac{\alpha }{2},n-1}\right)$

Hence the null hypothesis is rejected.

The predictor BachDeg% is significant.

Hence, the predictor BachDeg% significantly from zero at $\alpha =0.05$.