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2 years ago

A pizza parlor offers 4 different pizza toppings. How many different kinds of 2-topping pizzas are available?

2 answers:

6 0

Order does not matter so use "n choose k" formula is used to find number of unique combinations.

c=n!/(k!(n-k)!) where n is total possible choices and k is number of selections.

c=4!/(2!(4-2)!)

c=4!/(2!2!)

c=24/(2*2)

c=24/4

c=6

So there are 6 different two topping options when there are four different toppings to choose from.

Murrr4er [49]2 years ago

3 0

There are six different options

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$p_v = 2*P(t_{n-2} > |t_{calc}|)= 0.91$

So on this case for the significance level assumed $\alpha=0.05$ we see that $p_v >\alpha$ so then we can conclude that the result is NOT significant. And we don't have enough evidence to reject the null hypothesis.

So on this case is not appropiate say that :"the more we spend on advertising this product, the fewer units we sell" since the slope for this case is not significant.

Step-by-step explanation:

Let's suppose that we have the following linear model:

$y= \beta_o +\beta_1 X$

Where Y is the dependent variable and X the independent variable. $\beta_0$ represent the intercept and $\beta_1$ the slope.

In order to estimate the coefficients $\beta_0 ,\beta_1$ we can use least squares procedure.

If we are interested in analyze if we have a significant relationship between the dependent and the independent variable we can use the following system of hypothesis:

Null Hypothesis: $\beta_1 = 0$

Alternative hypothesis: $\beta_1 \neq 0$

Or in other words we want to check is our slope is significant (X have an effect in the Y variable )

In order to conduct this test we are assuming the following conditions:

a) We have linear relationship between Y and X

b) We have the same probability distribution for the variable Y with the same deviation for each value of the independent variable

c) We assume that the Y values are independent and the distribution of Y is normal

The significance level assumed on this case is $\alpha=0.05$

The standard error for the slope is given by this formula:

$SE_{\beta_1}=\frac{\sqrt{\frac{\sum (y_i -\hat y_i)^2}{n-2}}}{\sqrt{\sum (X_i -\bar X)^2}}$

Th degrees of freedom for a linear regression is given by $df=n-2$ since we need to estimate the value for the slope and the intercept.

In order to test the hypothesis the statistic is given by:

$t=\frac{\hat \beta_1}{SE_{\beta_1}}$

The p value on this case would be given by:

$p_v = 2*P(t_{n-2} > |t_{calc}|)= 0.91$

So on this case is not appropiate say that :"the more we spend on advertising this product, the fewer units we sell" since the slope for this case is not significant.

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