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77julia77 [94]
3 years ago
14

At many golf​ clubs, a teaching professional provides a free​ 10-minute lesson to new customers. A golf magazine reports that go

lf 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
Mathematics
1 answer:
Bezzdna [24]3 years ago
8 0

Answer:

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,\alpha as 0.10. This means that:

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

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A = \pi r^{2}

so for choice (A) area is given as 4.375 \times 10^{7}. So plug this in A place in formula and then solve for radius r

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diameter = 2 \times r = 2 \times 3731.76 = 7463.52 inches

Similarly for choice (E) area is 1.428 \times 10^{1} = 14.28
so again plug in formula

A = \pi r^{2}

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