Question

(1 point) A random sample of 100 observations from a population with standard deviation 25.11 yielded...

(1 point) A random sample of 100 observations from a population with standard deviation 25.11 yielded a sample mean of 94.2

1. Given that the null hypothesis is ?=90and the alternative hypothesis is ?>90 using ?=.05, find the following:
(a) Test statistic =

(b) P - value:

(c) The conclusion for this test is:
_____A. There is insufficient evidence to reject the null hypothesis
______B. Reject the null hypothesis
______C. None of the above

2. Given that the null hypothesis is ?=90

and the alternative hypothesis is ?≠90 using ?=.05, find the following:
(a) Test statistic =
(b) P - value:

(c) The conclusion for this test is:
_______A. There is insufficient evidence to reject the null hypothesis
_______B. Reject the null hypothesis
______C. None of the above

2. (1 point) 35 people are randomly selected and the accuracy of their wristwatches is checked, with positive errors representing watches that are ahead of the correct time and negative errors representing watches that are behind the correct time. The 35 values have a mean of 92sec and a standard deviation of 200sec. Use a 0.02 significance level to test the claim that the population of all watches has a mean of 0 sec.

The test statistic is_____

The P-Value is_______

The final conclustion is
____A. There is sufficient evidence to warrant rejection of the claim that the mean is equal to 0
____B. There is not sufficient evidence to warrant rejection of the claim that the mean is equal to 0

3. (1 point) Golf-course designers have become concerned that old courses are becoming obsolete since new technology has given golfers the ability to hit the ball so far. Designers, therefore, have proposed that new golf courses need to be built expecting that the average golfer can hit the ball more than 260 yards on average. Suppose a random sample of 151 golfers be chosen so that their mean driving distance is 264.3 yards, with a standard deviation of 45.8 Conduct a hypothesis test where ?0:?=260 and ?1:?>260 by computing the following:
(a) test statistic ______
(b) p-value ?= ________
(c) If this was a two-tailed test, then the p-value is ________

Homework Answers

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
1) The sample mean and standard deviation from a random sample of 21 observations from a...
1) The sample mean and standard deviation from a random sample of 21 observations from a normal population were computed as ?¯=25 and s = 8. Calculate the t statistic of the test required to determine whether there is enough evidence to infer at the 6% significance level that the population mean is greater than 21. Test Statistic = 2) Golf-course designers have become concerned that old courses are becoming obsolete since new technology has given golfers the ability to...
Golf-course designers have become concerned that old courses are becoming obsolete since new technology has given...
Golf-course designers have become concerned that old courses are becoming obsolete since new technology has given golfers the ability to hit the ball so far. Designers, therefore, have proposed that new golf courses need to be built expecting that the average golfer can hit the ball more than 260 yards on average. Suppose a random sample of 184 golfers be chosen so that their mean driving distance is 257.3 yards. The population standard deviation is 44. Use a 5% significance...
A random sample of 100 observations from a population with standard deviation 17.18 yielded a sample...
A random sample of 100 observations from a population with standard deviation 17.18 yielded a sample mean of 93.3 1. Given that the null hypothesis is μ=90 and the alternative hypothesis is μ>90 using α=.05, find the following: (a) Test statistic = (b) P - value: (c) The conclusion for this test is: A. There is insufficient evidence to reject the null hypothesis B. Reject the null hypothesis C. None of the above 2. Given that the null hypothesis is...
A random sample of 100 observations from a population with standard deviation 23.35 yielded a sample...
A random sample of 100 observations from a population with standard deviation 23.35 yielded a sample mean of 94.5. 1. Given that the null hypothesis is μ=90, and the alternative hypothesis is μ>90 and using α=.05, find the following: (a) Test statistic = (b) P - value: (c) The conclusion for this test is: A. Reject the null hypothesis B. There is insufficient evidence to reject the null hypothesis C. None of the above 2. Given that the null hypothesis...
A random sample of 100100 observations from a population with standard deviation 10.7610.76 yielded a sample...
A random sample of 100100 observations from a population with standard deviation 10.7610.76 yielded a sample mean of 91.891.8. 1. Given that the null hypothesis is μ=90μ=90 and the alternative hypothesis is μ>90μ>90 using α=.05α=.05, find the following: (a) Test statistic ==   (b)  P - value:    (c) The conclusion for this test is: A. There is insufficient evidence to reject the null hypothesis B. Reject the null hypothesis C. None of the above 2. Given that the null hypothesis is μ=90μ=90...
A random sample of 100100 observations from a population with standard deviation 12.6312.63 yielded a sample...
A random sample of 100100 observations from a population with standard deviation 12.6312.63 yielded a sample mean of 92.392.3. 1. Given that the null hypothesis is μ=90μ=90 and the alternative hypothesis is μ>90μ>90 using α=.05α=.05, find the following: (a) Test statistic == (b) P - value: (c) The conclusion for this test is: A. Reject the null hypothesis B. There is insufficient evidence to reject the null hypothesis C. None of the above 2. Given that the null hypothesis is...
Golf-course designers have become concerned that old courses are becoming obsolete since new technology has given...
Golf-course designers have become concerned that old courses are becoming obsolete since new technology has given golfers the ability to hit the ball so far. Designers, therefore, have proposed that new golf courses need to be built expecting that the average golfer can hit the ball more than 235 yards on average. Suppose a random sample of 143 golfers be chosen so that their mean driving distance is 232.5 yards. The population standard deviation is 47.6. Use a 5% significance...
A random sample of 100observations from a population with standard deviation 23.99 yielded a sample mean...
A random sample of 100observations from a population with standard deviation 23.99 yielded a sample mean of 94.1 1. Given that the null hypothesis is μ=90μ=90 and the alternative hypothesis is μ>90μ>90 using α=.05α=.05, find the following: (a) Test statistic == (b) P - value: (c) The conclusion for this test is: A. There is insufficient evidence to reject the null hypothesis B. Reject the null hypothesis C. None of the above 2. Given that the null hypothesis is μ=90μ=90...
Golf-course designers have become concerned that old courses are becoming obsolete since new technology has given...
Golf-course designers have become concerned that old courses are becoming obsolete since new technology has given golfers the ability to hit the ball so far. Designers, therefore, have proposed that new golf courses need to be built expecting that the average golfer can hit the ball more than 240 yards on average. Suppose a random sample of 113 golfers be chosen so that their mean driving distance is 237.6 yards. The population standard deviation is 41.8. Use a 5% significance...
45 people are randomly selected and the accuracy of their wristwatches is checked, with positive errors...
45 people are randomly selected and the accuracy of their wristwatches is checked, with positive errors representing watches that are ahead of the correct time and negative errors representing watches that are behind the correct time. The 45 values have a mean of 98sec and a standard deviation of 192sec. Use a 0.01 significance level to test the claim that the population of all watches has a mean of 0sec. The test statistic is The P-Value is The final conclustion...