Question 1: Calculating covariance and interpreting a negative covariance

a)

 

Cov (x, y) = -89/8-1

= -89/7

= -12.71

b). The covariance between years of experience and salary is -12.71. A negative covariance implies that the variables are inversely related (Stevens, 2012).

c)

= -12.71/2.45 x 5.24

= -1

A correlation coefficient of -1 is an indication that for every positive increase in years of experience there is a negative decrease of a fixed proportion in the salary (Cohen and Kohn, 2011).

d). A negative correlation implies that the relationship between years of experience and salary is negative 100% of the time (Koo and Li, 2016;Mukaka, 2012).

a).

H0: p = 0.1

H1:  p≠  0.1

b). We calculate the test statistic;

p= 81/900

= 0.09

Z^∗=p−p_o/(√p_o×(1−p_o)/n))

= 0.09 – 0.1 /√0.1 x0.9)/900))

= -0.01/√0.0004

= -0.01/0.02

Z^∗=- 0.5

We next ascertain the region of rejection for the level of significance and also the p-value for the test statistic;

From the Z-table and alpha value of 0.025;

Z = -2.81

Since the test statistic Z^∗ of – 0.5 falls outside the area of rejection (more than -2.81), the null hypothesis is accepted

With the p-value close to zero and being more than alpha of 0.05, we accept the null hypothesis. There is statistical evidence at 5% significance level to conclude that 10% of the users of a certain sinus drug experience drowsiness.

c).

CI = [X¯ − tα/2 σ √ n , X¯ + tα/2 σ √ n ]

SEX¯ = s √ n

=81/√900

= 2.7

= t_α/2,_n−1? = 2.045

CI = 81 – 2.045 X2.7, 81 + 2.045 X 2.7

= 75.48, 86.52)

= (75, 87)

d). If the confidence interval has the value claimed by the null hypothesis, then the findings are close to the claimed value, hence null hypothesis is not rejected (Javanmard and Montanari, 2014;Greenland et al., 2016).

On the other hand if the interval doesn’t have the claimed value by the H_O, then the findings vary significantly from the claimed value, thus null hypothesis is rejected (Murphy, Myors, and Wolach, 2014; Siegmund, 2013).

a).

student	Rent (x)	X-Xbar	(X-Xbar)sqd
1	730.00	5.33	28.44
2	690.00	–       34.67	1,201.78
3	560.00	–     164.67	27,115.11
4	600.00	–     124.67	15,541.78
5	730.00	5.33	28.44
6	730.00	5.33	28.44
7	1,030.00	305.33	93,228.44
8	740.00	15.33	235.11
9	620.00	–     104.67	10,955.11
10	800.00	75.33	5,675.11
11	730.00	5.33	28.44
12	740.00	15.33	235.11
13	650.00	–       74.67	5,575.11
14	760.00	35.33	1,248.44
15	820.00	95.33	9,088.44
16	930.00	205.33	42,161.78
17	620.00	–     104.67	10,955.11
18	660.00	–       64.67	4,181.78
19	690.00	–       34.67	1,201.78
20	840.00	115.33	13,301.78
21	700.00	–       24.67	608.44
22	720.00	–          4.67	21.78
23	850.00	125.33	15,708.44
24	710.00	–       14.67	215.11
25	720.00	–          4.67	21.78
26	570.00	–     154.67	23,921.78
27	670.00	–       54.67	2,988.44
28	930.00	205.33	42,161.78
29	500.00	–     224.67	50,475.11
30	700.00	–       24.67	608.44
TOTAL	21,740.00		378,746.67

Mean = Sum of observations/Total number of observations

 = 21,740/30

= 724.67

Median  

Median = (n/2)^th(n/2)th observation + (n/2 + 1)th observation

730, 500, 560, 570,600,620, 620,650,660,670,690,690,700,700,710,720,720,730,730,740,760,820,840,850,930,930,730,740, 800, 1030

Median  = 720

Mode = Value with highest frequency

Mode = 730

b). Yes, they are pretty close to each other

c).

= 378,746.67/29

=114.28

1. Yes, there is only one outlier data value, that is 1,030. This data value is far outside the normal range of all others which is hundreds
2. Yes, because the measures of central tendency agree and also there are no significant outliers data values.

a).

P (A and O)

= P (0.6 x 0.8)

= 0.48

b).

P (a package delivery on time)

=P (0.6 x0.8) + (0.3 x 0.6) + (0.1 x 0.4)

= 0.48 + 0.18 + 0.04

=0.7

c).

P( a package delivered on time) = 0.7

If was source A = 0.7 x 0.6

=0.42

d).

Prob package late

=P (service B delivery late)

= (0.3 x (1-0.6)

= 0.12

e).

P (service C delivery late)

= 0.3 x (1-0.4)

=0.18

References

Cohen, M.R. and Kohn, A., 2011. Measuring and interpreting neuronal correlations. Nature neuroscience, 14(7), p.811.

Greenland, S., Senn, S.J., Rothman, K.J., Carlin, J.B., Poole, C., Goodman, S.N. and Altman, D.G., 2016. Statistical tests, P values, confidence intervals, and power: a guide to misinterpretations. European journal of epidemiology, 31(4), pp.337-350.

Javanmard, A. and Montanari, A., 2014. Confidence intervals and hypothesis testing for high-dimensional regression. The Journal of Machine Learning Research, 15(1), pp.2869-2909.

Koo, T.K. and Li, M.Y., 2016. A guideline of selecting and reporting intraclass correlation coefficients for reliability research. Journal of chiropractic medicine, 15(2), pp.155-163.

Mukaka, M.M., 2012. A guide to appropriate use of correlation coefficient in medical research. Malawi Medical Journal, 24(3), pp.69-71.

Murphy, K.R., Myors, B. and Wolach, A., 2014. Statistical power analysis: A simple and general model for traditional and modern hypothesis tests. Routledge.

Siegmund, D., 2013. Sequential analysis: tests and confidence intervals. Springer Science & Business Media.

Stevens, J.P., 2012. Analysis of covariance. In Applied Multivariate Statistics for the Social Sciences, Fifth Edition (pp. 299-326). Routledge.