Question

In general, high school and college students are the most pathologically sleep-deprived segment of the population. Their alertness during the day is on par with that of untreated narcoleptics and those with untreated sleep apnea. Not surprisingly, teens are also 71 percent more likely to drive drowsy and/or fall asleep at the wheel compared to other age groups. (Males under the age of twenty-six are particularly at risk.)

The accompanying data set represents the number of hours 25 college students at a small college in the northeastern United States slept and is from a random sample. Enter this data into C1 of Minitab Express.

6 9 7 7 6 7 7 5 8 6 6 6 8 8 8 5 4 6 7 8 5 8 7 6 7

For the analyses that follow, we shall use

·         90%, 95%, and 99% as the confidence levels for the confidence interval.

·      5% as the level of significance ( ) for the hypothesis test.

·         7 hours sleep as the null hypothesis (according to The Sleep Foundation).

a.    List the three (3) assumptions for a valid confidence interval and hypothesis test. Provide an explanation as to whether or not each one is met - more than just a simple “yes” or “no” – and refer to the boxplot and normal probability plot, as necessary, in your assessment.

b.    What degrees of freedom will you use for the t distribution? Show your calculation.

(Hint: degrees of freedom is n-1.)

Answer 1

a.    List the three (3) assumptions for a valid confidence interval and hypothesis test. Provide an explanation as to whether or not each one is met - more than just a simple “yes” or “no” – and refer to the boxplot and normal probability plot, as necessary, in your assessment.

Here, we have to check the assumption of normality by using box plot and normal probability plot. Box plot and normal probability plot by using Minitab are given as below:

Boxplot Sleep in hours

From this boxplot, it is observed that there is no significant outlier detected and there is no high skewed nature of the distribution is observed. Mean and median are approximately near to each other.

Distribution Function Analysis

Normal Dist. Parameter Estimates (ML)

Variable: Sleep in hour

Mean    6.68  

StDev   1.19063

Goodness of Fit

Anderson-Darling (adjusted) = 1.186

Percentile Estimates

                     95% CI       95% CI    

                     Approximate Approximate

Percent Percentile Lower Limit Upper Limit

1       3.91018     3.01171       4.8087   

2       4.23474     3.41182       5.0577   

3       4.44067     3.66408       5.2173   

4       4.59558     3.85286       5.3383   

5       4.72159     4.00570       5.4375   

6       4.82884     4.13522       5.5225   

7       4.92288     4.24832       5.5974   

8       5.00708     4.34917       5.6650   

9       5.08366     4.44053       5.7268   

10       5.15415     4.52430       5.7840   

20       5.67794     5.13483       6.2211   

30       6.05563     5.55786       6.5534   

40       6.37836     5.90421       6.8525   

50       6.68000     6.21328       7.1467   

60       6.98164     6.50749       7.4558   

70       7.30437     6.80660       7.8021   

80       7.68206     7.13895       8.2252   

90       8.20585     7.57601       8.8357   

91       8.27634     7.63322       8.9195   

92       8.35292     7.69501       9.0108   

93       8.43712     7.76256       9.1117   

94       8.53116     7.83754       9.2248   

95       8.63841     7.92252       9.3543   

96       8.76442     8.02170       9.5071   

97       8.91933     8.14274       9.6959   

98       9.12526     8.30233       9.9482   

99       9.44982     8.55135      10.3483

Prob. Plot for Sleep in hour

From this normal probability plot, it is observed that the given data follows an approximate normal distribution because all points are close to line.

Part b

What degrees of freedom will you use for the t distribution? Show your calculation.

We are given sample size = n = 25

Degrees of freedom = df = n – 1 = 25 – 1 = 24

df = 24