Question

In: Statistics and Probability

Question 5 (1 point) The owner of a local golf course wants to determine the average...

Question 5 (1 point)

The owner of a local golf course wants to determine the average age of the golfers that play on the course in relation to the average age in the area. According to the most recent census, the town has an average age of 48.93. In a random sample of 24 golfers that visited his course, the sample mean was 48.1 and the standard deviation was 5.999. Using this information, the owner calculated the confidence interval of (44.66, 51.54) with a confidence level of 99%. Which of the following statements is the best conclusion?

Question 5 options:

1)

We are 99% confident that the average age of all golfers that play on the golf course is less than 48.93

2)

The percentage of golfers with an age greater than 48.93 is 99%.

3)

We cannot determine the proper interpretation based on the information given.

4)

The average age of all golfers does not significantly differ from 48.93.

5)

We are 99% confident that the average age of all golfers that play on the golf course is greater than 48.93

Question 6 (1 point)

As an avid golfer, you want to estimate your average score for 18 holes of golf. Suppose you know that the standard deviation of your score is 15.272 strokes and you want to find a sample mean that is within 5.321 strokes of your true average for all rounds of golf with 90% confidence. How many rounds would you need to play to determine this?

Question 6 options:

1)

33

2)

28

3)

22

4)

We do not have enough information to answer this question since we were not given the sample mean.

5)

23

Question 7 (1 point)

You are in the market for a new car. You want to check whether there is a significant difference between the fuel economy of mid-size domestic cars and mid-size import cars. You sample 16 domestic car makes and find an average fuel economy of 32.487 MPG with a standard deviation of 4.082 MPG. For imports, you sample 17 cars and find an average MPG of 32.065 MPG with a standard deviation of 6.763. If a 95% confidence interval is calculated to estimate the difference between the average fuel economy of domestic and import mid-size cars, what is the margin of error? Assume both population standard deviations are equal.

Question 7 options:

1)

3.998

2)

3.842

3)

2.03951345

4)

1.96016429

5)

3.323

Question 8 (1 point)

The owner of a local golf course wants to estimate the difference between the average ages of males and females that play on the golf course. He randomly samples 12 men and 13 women that play on his course. He finds the average age of the men to be 49.38 with a standard deviation of 5.502. The average age of the women was 36.37 with a standard deviation of 12.14. Construct a 99% confidence interval to estimate the difference of (average age of men - average age of women). Assume the population standard deviations are the same.

Question 8 options:

1)

(3.15, 22.87)

2)

We only have the sample means, we need to know the population means in order to calculate a confidence interval.

3)

(9.18, 16.84)

4)

(2.27, 23.75)

5)

(-89.68, 115.7)

Solutions

Expert Solution

5. Since the confidence interval contains 48.93, so at 99% level of confidence, we can conclude that mean is not significantly different from 48.93.

ans-> 4) The average age of all golfers does not significantly differ from 48.93.

6. To find sample size (n) such that

Margin of error = 5.321

ans-> 5) 23

7.

ans-> 1) 3.998

8.

ans-> 4) (2.27, 23.75)


Related Solutions

The owner of a local golf course wants to determine the average age of the golfers...
The owner of a local golf course wants to determine the average age of the golfers that play on the course in relation to the average age in the area. According to the most recent census, the town has an average age of 63.248. In a random sample of 28 golfers that visited his course, the sample mean was 48.356 and the standard deviation was 6.8424. Using this information, the owner calculated the confidence interval of (45.703, 51.009) with a...
The owner of a local golf course wants to examine the difference between the average ages...
The owner of a local golf course wants to examine the difference between the average ages of males and females that play on the golf course. Specifically, he wants to test if the average age of males is greater than the average age of females. If the owner conducts a hypothesis test for two independent samples and calculates a p-value of 0.5441, what is the appropriate conclusion? Label males as group 1 and females as group 2. 1) We did...
Question 5 (1 point) A student at a university wants to determine if the proportion of...
Question 5 (1 point) A student at a university wants to determine if the proportion of students that use iPhones is less than 0.46. The hypotheses for this scenario are as follows. Null Hypothesis: p ≥ 0.46, Alternative Hypothesis: p < 0.46. If the student takes a random sample of students and calculates a p-value of 0.8906 based on the data, what is the appropriate conclusion? Conclude at the 5% level of significance. Question 5 options: 1) The proportion of...
Question 2 (1 point) The owner of a local phone store wanted to determine how much...
Question 2 (1 point) The owner of a local phone store wanted to determine how much customers are willing to spend on the purchase of a new phone. In a random sample of 15 phones purchased that day, the sample mean was $302.071 and the standard deviation was $27.9396. Calculate a 99% confidence interval to estimate the average price customers are willing to pay per phone. Question 2 options: 1) ( -280.596 , 323.546 ) 2) ( 280.596 , 323.546...
Question 1 A local realtor wants to determine the average value for single-family houses in a...
Question 1 A local realtor wants to determine the average value for single-family houses in a community.A random sample of 50 houses is selected. The sample mean value is found to be $220,000 with a sample standard deviation of $40,000. a) Assuming that the values of all single-family houses in the community is normally distributed, compute a 90% confidence interval estimate for μ, where μ is the population mean value for all single-family houses in the community. b) Test the...
The owner of a local supermarket wants to estimate the difference between the average number of...
The owner of a local supermarket wants to estimate the difference between the average number of gallons of milk sold per day on weekdays and weekends. The owner samples 8 weekdays and finds an average of 218.91 gallons of milk sold on those days with a standard deviation of 33.376. 10 (total) Saturdays and Sundays are sampled and the average number of gallons sold is 377.74 with a standard deviation of 49.365. Construct a 99% confidence interval to estimate the...
The owner of a factory wants to determine a confidence interval for the average wages at...
The owner of a factory wants to determine a confidence interval for the average wages at his factory. The population average wages is unknown. However, the owner was able to take a random sample of 64 workers. He found that the average salary for this sample of workers is $50,000. The population standard deviation is $3000. He calculated a 99% confidence interval for the population average wages at the factory. Which of the following choices is correct? Assume the distribution...
Norb and Gary are entered in a local golf tournament. Both have played the local course...
Norb and Gary are entered in a local golf tournament. Both have played the local course many times. Their scores are random variables with the following means and standard deviations. Norb, x1: μ1 = 115; σ1 = 8 Gary, x2: μ2 = 100; σ2 = 18 In the tournament, Norb and Gary are not playing together, and we will assume their scores vary independently of each other. (a) The difference between their scores is W = x1 − x2. Compute...
A local electronics firm wants to determine their average daily sales (in dollars.) A sample of...
A local electronics firm wants to determine their average daily sales (in dollars.) A sample of the sales for 36 days revealed an average sales of $141,000. Assume that the standard deviation of the population is known to be $12,000. Then, the 95% margin of error will be . . . Provide a 95% confidence interval estimate for the true average daily sales. $137,080 to $144,920 $139,500 to $143,000 $132,800 to 139,200 $136,200 to $142,200 $140,600 to $143,200 $135,600 to...
A golf instructor wants to determine if the drive distance is different for three golf ball...
A golf instructor wants to determine if the drive distance is different for three golf ball brands. The instructor has several of his students drive the different brands and collects the following data. Golf Ball Brands Golfer A B C Brad 275 350 295 Will 300 525 380 Huang 245 286 300 Singh 220 256 286 Woods 400 415 475 Fisher 356 385 396 The following information is provided: SSTR = 16520; SSE = 20335; SSTotal = 114520. Please construct...
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT