Question

In the fall of 2008, the Cleveland Plain Dealer identified 30 players to watch during the upcoming girls' high school basketball season. Although these 30 players are not a random sample of female high school basketball players, for this problem, consider their heights to be a random sample of heights of female high school basketball players. In Spring, 2006, 29 male students in a Psychology 101 course reported their heights. Although these 29 students are not a random sample of male college students, consider them to be a random sample for the purpose of this problem. The stemplot below shows the distributions of heights in the two samples and provides sample means and standard deviations. MALE COLLEGE STUDENTS Inches (N = 29) FEMALE HIGH SCHOOL BASKETBALL PLAYERS (N = 30) 5 6 6 5 7666 988 11110000000 33222 44447 7 3 455 6667 888999 00011 222233 44444 7 70.5 MEAN69.6 2.9 SD 3.3
1) Refer carefully to the question that is being asked.
(i) Specify the null and alternative hypotheses that are being tested and
(ii) define the us used in your hypotheses. (These hypotheses should be about two population means.)
2) You will calculate at statistic. (0) Using Option 2 (Moore et al., The Basic Practice of Statistics, p. 484) to determine degrees of freedom, indicate how many degrees of freedom are involved, and (ii) whether the P value will be one-sided or two-sided.
3) (i) Calculate the t statistic and (ii) find its P value.
4) With a = .01, decide whether to reject the null hypothesis. (At least say "Reject H," or Retain H,"; additional explanation of your answer is optional.)

Answer 1

Answer:

Check the explanation

Step-by-step explanation:

a ) let u(1) is the population at a mean rate for the college students  who are of the male gender.

u(2) is the population mean for female college students

H0:u(1) = u(2)

Ha:u(1) \neq u(2 )

2 ) using minitab>stat>basic stat>two sample t

we have

Two-Sample T-Test and CI

Sample N Mean StDev SE Mean

1 29 70.50 2.90 0.54

2 30 69.60 3.30 0.60

Difference = μ (1) - μ (2)

Estimate for difference: 0.900

95% CI for difference: (-0.722, 2.522)

T-Test of difference = 0 (vs ≠): T-Value = 1.11 P-Value = 0.271 DF = 57

Both use Pooled StDev = 3.1099

degree of freedom = 57

here p value will be two sided

3 ) using the output

t value = 1.11

p value = 0.271

4 ) since p value is greater than 001 so we will accept the null hypothesis .

we will conclude that average of male and female college basketball players are same