91Ó°ÊÓ

It is given that y is the dependent variable (here, cost per use). The test to be conducted at $\mathrm{Î\pm }=.10$.

Dummy variables indicate the presence or absence of the categorical effect which may shift the outcome. Dummy variable can take only the value 0 or 1.

a.

For qualitative regression model,(k-1) variables are introduced in the model where k = no. of levels. Since in this question there are only 2 levels, 1 qualitative variable x₁ is introduced in the model where x₁ = 1 when repellent lotion is based, 0 otherwise.

b. From the excel output, the regression equation becomes $E\left(y\right)=0.7775+0.109167x_1$.

For the anova table one need to calculate the mean of the independent variable and then calculate the SSR, SSE AND SST, after that one need to calculate the degrees of freedom and the mean squares and then F.

The SSR is calculated by using$n\mathrm{Î\pounds }{\left(X_j--{\stackrel{¯}{x}}_j..\right)}^2$, and the SSE is calculated by squaring the each term and adding them all. The SST is the sum of SSR and SSE. The MS regression is calculated by dividing SST by degrees of regression and similarly the MS residual is calculated by dividing SSE by degrees of residual and F is calculated by dividing MS regression by MS residual.

The coefficients of x is calculated by using this formula:$\frac{n\left(∑xy\right)-\left(∑x\right)\left(∑y\right)}{n\left(∑x_2\right)-\left(∑x_2\right)}$whereas the coefficient of intercept is calculated by $\frac{\left(∑y\right)\left(∑x_2\right)-\left(∑x\right)\left(∑xy\right)}{n\left(∑x_2\right)-\left(∑x_2\right)}$.

The standard error is calculated by dividing the standard deviation by the sample size's square root.

c.

The null and alternate hypothesis for whether repellent type is a useful predictor of cost per use (y) can be written as$H_0\text{:}{\mathrm{β}}_1=0$ against $H_a\text{:}{\mathrm{β}}_1≠0$.

d.

$$\begin{gathered} H_0\text{:}{\mathrm{β}}_1=0 \\ {H}_a\text{:}{\mathrm{β}}_1≠0 \end{gathered}$$

Here, t-test statistic

.$$\begin{gathered} t=\frac{\stackrel{Áåž}{{\mathrm{β}}_1}}{S\stackrel{Áåž}{{\mathrm{β}}_1}} \\ =\frac{0.109167}{0.454457} \\ =0.24021 \end{gathered}$$

Value of$t_{0.05,13}$is 1.771.

H₀ is rejected if t statistic$>t_{0.05,13}$.

For$\mathrm{Î\pm }=0.05$,since$<t_{0.05,13}$.

Not sufficient evidence to reject H₀ at 95% confidence interval.

Therefore,${\mathbf{β}}_{\mathbf{3}}\mathbf{=}\mathbf{0}$meaning thatrole="math" localid="1660794217559" ${\mathbf{β}}_{\mathbf{3}}$is statistically significant for the model.

e.

For the anova table one need to calculate the mean if the independent variable and then calculate the SSR, SSE AND SST, after that onee need to calculate the degrees of freedom and the mean squares and the F.

The SSR is calculated by using$n\mathrm{Î\pounds }(X_i-\stackrel{Áåœ}{X_j})$, and the SSE is calculated by squaring the each term and adding them all. The SST is the sum of SSR and SSE. The MS regression is calculated by dividing SST by degrees of regression and similarly the MS residual is calculated by dividing SSE by degrees of residual and F is calculated by dividing MS regression by MS residual.

The coefficients of x is calculated by using this formula: $\frac{n\left(∑xy\right)-\left(∑x\right)\left(∑y\right)}{n\left(∑x_2\right)-\left(∑x_2\right)}$ whereas the coefficient of intercept is calculated by $\frac{\left(∑y\right)\left(∑x2\right)-\left(∑x\right)\left(∑xy\right)}{n\left(∑x_2\right)-\left(∑x_2\right)}$.

The standard error is calculated bydividing the standard deviation by the sample size's square root.

The R-squared is calculated as$R_2=SSyy--\frac{SSE}{SSyy}$.

The excel output is

The regression equation for maximum hours of protection as a function of repellent type is$E\left(y\right)=7.5625-1.64583x_1$

To test the significance of${\mathrm{β}}_1$

$$\begin{gathered} H_0\text{:}{\mathrm{β}}_1=0 \\ {H}_a\text{:}{\mathrm{β}}_1≠0 \end{gathered}$$

Here, t-test statistic

$$\begin{gathered} t=\frac{\stackrel{Áåž}{{\mathrm{β}}_1}}{S\stackrel{Áåž}{{\mathrm{β}}_1}} \\ =\frac{-1.64583}{3.573625} \\ =-0.46054 \end{gathered}$$

Value of $t_{0.05,13}$is 1.771

H₀ is rejected if t statistic $>t_{0.05,13}$

For,since$>t_{0.05,13}$.

Not sufficient evidence to reject H₀ at95%confidence interval.

Therefore,${\mathbf{β}}_{\mathbf{3}}\mathbf{=}\mathbf{0}$meaning that${\mathbf{β}}_{\mathbf{3}}$ is statistically significant for the model.