Lat year, the mean dollar spent per transaction for online purchases is $206. the population standard deviation is not known. because of increased use of on-line purchasing, vista's vice president of electronic marketing believes that purchasing has increased. he randomly selects 100 customer accounts. the results of the sample found that customers spent a mean dollar amount of $214 on purchases with s= $55. (use alpha 5%) a) state the null and alternative hypothesis in symbols and in words. (3pts) b) calculate the expected results for the hypothesis test sampling distribution, assuming the null hypothesis is true. (i.e., name and graph of sampling distribution, mean and standard error) (3pts) c) identify the standard distribution that best approximates the sampling distribution. (1pts) d) formulate the decision rule(use either critical test scores or p-values) (3pts) e) determine the statistical results: test statistic, critical test statistic and p-value ( 6pts) f) determine the conclusion in terms of the null and alternative hypotheses. do the sample results indicate that the vice president's claim is supported at alpha = 5%? (4pts)

The question is incomplete! Complete question along with answer and step by step explanation is provided below.

Question:

Miguel is a golfer, and he plays on the same course each week. The following table shows the probability distribution for his score on one particular hole, known as the Water Hole.  

Score 3 4 5 6 7

Probability 0.15 0.40 0.25 0.15 0.05

Let the random variable X represent Miguel’s score on the Water Hole. In golf, lower scores are better.

(a) Suppose one of Miguel’s scores from the Water Hole is selected at random. What is the probability that Miguel’s score on the Water Hole is at most 5 ? Show your work.

(b) Calculate and interpret the expected value of X . Show your work.

A potential issue with the long hit is that the ball might land in the water, which is not a good outcome. Miguel thinks that if the long hit is successful, his expected value improves to 4.2. However, if the long hit fails and the ball lands in the water, his expected value would be worse and increases to 5.4.

c) Suppose the probability of a successful long hit is 0.4. Which approach, the short hit or long hit, is better in terms of improving the expected value of the score?

(d) Let p represent the probability of a successful long hit. What values of p will make the long hit better than the short hit in terms of improving the expected value of the score? Explain your reasoning.

a) 80%

b) 4.55

c) 4.92

d) P > 0.7083

Step-by-step explanation:

Score  |   Probability

3          |      0.15

4          |      0.40

5          |      0.25

6          |      0.15

7          |      0.05

Let the random variable X represents Miguel’s score on the Water Hole.

a) What is the probability that Miguel’s score on the Water Hole is at most 5 ?

At most 5 means scores which are equal or less than 5

P(at most 5) = P(X ≤ 5) = P(X = 3) + P(X = 4) + P(X = 5)

P(X ≤ 5) = 0.15 + 0.40 + 0.25

P(X ≤ 5) = 0.80

P(X ≤ 5) = 80%

Therefore, there is 80% chance that Miguel’s score on the Water Hole is at most 5.

(b) Calculate and interpret the expected value of X.

The expected value of random variable X is given by

E(X) = X₃P₃ + X₄P₄ + X₅P₅ + X₆P₆ + X₇P₇

E(X) = 3*0.15 + 4*0.40 + 5*0.25 + 6*0.15 + 7*0.05

E(X) = 0.45 + 1.6 + 1.25 + 0.9 + 0.35

E(X) = 4.55

Therefore, the expected value of 4.55 represents the average score of Miguel.

c) Suppose the probability of a successful long hit is 0.4. Which approach, the short hit or long hit, is better in terms of improving the expected value of the score?

The probability of a successful long hit is given by

P(Successful) = 0.40

The probability of a unsuccessful long hit is given by

P(Unsuccessful) = 1 - P(Successful)

P(Unsuccessful) = 1 - 0.40

P(Unsuccessful) = 0.60

The expected value of successful long hit is given by

E(Successful) = 4.2

The expected value of Unsuccessful long hit is given by

E(Unsuccessful) = 5.4

So, the expected value of long hit is,

E(long hit) = P(Successful)*E(Successful) + P(Unsuccessful)*E(Unsuccessful)

E(long hit) = 0.40*4.2 + 0.60*5.4

E(long hit) = 1.68 + 3.24

E(long hit) = 4.92

Since the expected value of long hit is 4.92 which is greater than the value of short hit obtained in part b that is 4.55, therefore, it is better to go for short hit rather than for long hit. (Note: lower expected score is better)

d) Let p represent the probability of a successful long hit. What values of p will make the long hit better than the short hit in terms of improving the expected value of the score?

The expected value of long hit is given by

E(long hit) = P(Successful)*E(Successful) + P(Unsuccessful)*E(Unsuccessful)

E(long hit) = P*4.2 + (1 - P)*5.4

We want to find the probability P that will make the long hit better than short hit

P*4.2 + (1 - P)*5.4 < 4.55

4.2P + 5.4 - 5.4P < 4.55

-1.2P + 5.4 < 4.55

-1.2P < -0.85

multiply both sides by -1

1.2P > 0.85

P > 0.85/1.2

P > 0.7083

Therefore, the probability of long hit must be greater than 0.7083 that will make the long hit better than the short hit in terms of improving the expected value of the score.