5:56 PM webwork.gonzaga.edu AT&T; 36% Prev Up Next (1 pt) The time needed for co

Question

5:56 PM webwork.gonzaga.edu AT&T; 36% Prev Up Next (1 pt) The time needed for college students to complete a certain paper-and-pencil maze follows a normal distribution with a mean of 30 seconds and a standard deviation of 2.2 seconds. You wish to see if the mean time is changed by vigorous exercise, so you have a group of 25 college students exercise vigorously for 30 minutes and then complete the maze. It takes them an average of x 27.3 seconds to complete the maze. Use this information to test the hypotheses Ho : = 30 Ha : 30 Conduct a test using a significance level of -0.01 (a) The test statistic -6.13636363 (b) The positive critical value, z*- (c) The final conclusion is A. There is not sufficient evidence to reject the null hypothesis that = 30. O B. We can reject the null hypothesis that = 30 and accept that 30.

Explanation / Answer

Solution:

Test Hypothesis:
H0: = 30
HA: 30
Sample mean x = 27.3 seconds
Standard deviation = 2.2 seconds

sample size n = 25
Standard error of mean = / n
Standard error of mean = 2.2 / 25
SE = 2.2/5
Standard error of mean = 0.44

t= (x - ) / SE
t = (27.3-30) / 0.44
t = -2.7/ 0.44

t(calculated) = -6.1364

level of significance = 0.01

degrees of freedom n-1 = 25-1 = 24

t tabulated for 0.01 level of significance and 24 degrees of freedom

is t tab = 2.7969
calculated value of t=|t|=|-6.1364|= 6.1364
table value of t = 2.7969
t cal > t tab
since calculated value of t doesn't fall in the acceptance region.
null hypothesis is not accepted.
reject the null hypothesis.

a) The test statistic = -6.1364
b) The positive critical value z* = 2.7969
c) The final conclusion is
B. We can reject the null hypothesis that = 30 and accept that 30.

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