At many golf clubs, a teaching professional provides a free 10-minute lesson to new customers. A golf magazine reports that golf facilities that provide these free lessons gain, on average, $2 comma 1002,100 in green fees, lessons, or equipment expenditures. A teaching professional believes that the average gain is notis not $2 comma 1002,100. Complete parts a through c below. a. In order to support the claim made by the teaching professional, what null and alternative hypotheses should you test? Upper H 0H0: muμ equals= $2 comma 1002,100 Upper H Subscript aHa: muμ not equals≠ $2 comma 1002,100 b. Suppose you select alphaαequals=0.100.10. Interpret this value in the words of the problem. The probablility that the null hypothesis is rejected when the average gain is less than $2 comma 1002,100 is

Given:

Mean, u = 2100

A golf magazine reports the mean gain to be $2100, while the teaching professional believes the average gain is not $2100.

Here the null and alternative hypotheses would be:

Null hypothesis:

H0: u = 2100

Alternative hypothesis:

Ha: u ≠ 2100

b) Here, given the level of significance, as 0.10. This means that:

The probability that the null hypothesis H0 is rejected when average gain is $2100 is 0.10

solution:

→c= a b , that is product of a and b is possible only when when a is m × n matrix and b is n × p matrix.

that is, number of columns of a= number of rows of b.

so, size of matrix , c= a b is , if a = (m × n)matrix

b= (n × p)matrix

then a b = [ (m × n)matrix] ×[   (n × p)matrix ]

→→c= a b= [m × p ] matrix

→→size of product a b will be equal to = number of rows of a × number of columns of b