A newly hired basketball coach promised a high-paced attack that will put mare p

Question

A newly hired basketball coach promised a high-paced attack that will put mare points on the board than the team's previously tepid offense historically managed. After a few months, the team owner looks at the data to test the coach's claim. He takes a sample of 36 of the team's games under the new coach and finds that they score an average of 101 points with a standard deviation of 6 points. Over the past 10 years, the team had averaged 99 points. What is (are) the appropriate critical value(s) to test the new coach's claim at the 1% significance level? A -2.438 B. 2.438 C. 2.326 D. -2.438 and 2.438 A university is interested in promoting graduates of its honors program by establishing that the mean GPA of these graduates exceeds 3.45. A sample of 36 honors students is taken and is found to have a mean GPA equal to 3.60. The population standard deviation is assumed to equal 0.40. Given that the H_0: mu lessthanorequalto 3.45 and the H_: mu > 3.45. (Place both answers on the line for 27) The value of the test statistic is _____. A. z = 2.25 B. z = -2.25 C. z = 1.5 D. z = -1.5 At the 5% significance level, the decision is to _____. (use p-value or critical value approach) A. Reject the null hypothesis B. Not reject the null hypothesis If the p-value for a hypothesis test is 030 and the chosen level of significance is a = .10, then the correct conclusion is to _______. A. Reject the null hypothesis B. Not reject the null hypothesis Which of the following identifies the range for a correlation coefficient? A. Any value between and including -1 and 1 B. Any value between 0 and 1 C. Any value less than 1 D. Any value greater than 0 A correlation coefficient r = .30 could indicate a: A. Very strong positive linear relationship. B. Very strong negative linear relationship. C. Very weak positive linear relationship. D. Very weak negative linear relationship.

Explanation / Answer

Question 26

Here, we have to find the critical values for the given test.

We are given

Sample mean = 101

Sample standard deviation = 6

We have to use t test for the population mean.

Level of significance = alpha = 0.01

Sample size = n = 36

Degrees of freedom = n – 1 = 36 – 1 = 35

By using t-table,

Critical value = 2.438

Question 27.a.

We are given

Sample size = n = 36

Sample mean = Xbar = 3.60

Population standard deviation = = 0.40

Hypothesized mean = µ = 3.45

Test is upper tailed.

Test statistic = Z = (Xbar - µ) / [/sqrt(n)]

Test statistic = Z = (3.60 – 3.45) / [0.40/sqrt(36)]

Test statistic = Z = 2.25

Z = 2.25

Question 27.b.

We are given,

Level of significance = alpha = 0.05 or 5%

P-value = 0.0122

P-value < Alpha value

So, we reject the null hypothesis

Reject the null hypothesis

Question 28

We are given p-value = 0.30

Alpha = 0.10

P-value > Alpha value

So, we do not reject the null hypothesis

Not reject the null hypothesis

Decision rule:

We reject the null hypothesis if p-value is less than the given level of significance or alpha value and we do not reject the null hypothesis if the p-value is greater than the given level of significance or alpha value.

Question 29

Answer:

Any value between and including -1 and 1

Question 30

Answer:

Very weak positive linear relationship

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