Question

Hello experts, any help with the given problems would be greatly appreciated.

1. Give and interpret the p-value at a=.05 for the hypothesis test where we wanted to see if the average hours of sleep is less than 8. Be sure to construct the Null and Alternative Hypothesis first.
2. If you use a=.01, does your interpretation in question 4 change? Explain your reasoning.
3. A student success coach has been told that the mean drive distance to class is 56 miles. Use our class data for Drive distance to test this claim – Use both a Rejection Region test and a P value test (level of significance of .05).

Drive (miles)	Sleep (hours)
36	7
20	7
88	5
6	7
71	6
42	8
76	7
63	8
36	5
63	8
38	8
28	4
55	8
33	8
40	6
80	8
86	8
83	8
4	7
39	10
25	6
25	7
54	8
54	5
81	8
73	7
29	7
76	4
78	9
77	8
42	7
36	7
71	8
94	8
6	10

Answer 1

Give and interpret the p-value at a=.05 for the hypothesis test where we wanted to see if the average hours of sleep is less than 8.

GIVEN:

I have used excel function to calculate mean and standard deviation of given data for sleep.

Sample size

Sample mean

Sample standard deviation

HYPOTHESIS:

(That is, the average hours of sleep is not significantly different from .)

(That is, the average hours of sleep is less than .)

LEVEL OF SIGNIFICANCE:

TEST STATISTIC:

P VALUE:

The left tailed p value for t statistic (-3.36) is .

DECISION RULE:

CONCLUSION:

Since the calculated p value (0.00039) is less than the significance level , we reject null hypothesis and conclude that the average hours of sleep is less than .

If we use ,

P VALUE:

The left tailed p value for t statistic (-3.36) is .

DECISION RULE:

CONCLUSION:

Since the calculated p value (0.00039) is less than the significance level , we reject null hypothesis and conclude that the average hours of sleep is less than .

A student success coach has been told that the mean drive distance to class is 56 miles.

GIVEN:

I have used excel function to calculate mean and standard deviation of given data for Drive.

Sample size

Sample mean

Sample standard deviation

HYPOTHESIS:

(That is, the mean drive distance to class is not significantly different from 56 miles.)

(That is, the mean drive distance to class is significantly different from 56 miles.)

LEVEL OF SIGNIFICANCE:

TEST STATISTIC:

CRITICAL VALUE:

The two tailed t critical value with degrees of freedom at significance level is .

DECISION RULE:

CONCLUSION:

Since the calculated t statistic value (-0.995) is greater than the t critical value (-2.032), we fail to reject null hypothesis and conclude that the mean drive distance to class is not significantly different from 56 miles.

P VALUE:

The two tailed p value for the given hypothesis test is .

DECISION RULE:

CONCLUSION:

Since the calculated p value (0.3267) is greater than the significance level , we fail to reject null hypothesis and conclude that the mean drive distance to class is not significantly different from 56 miles.