A sample of eight observations of variables x (years of experience) and y (salary in $1,000s) is shown below:

x |
5 |
3 |
7 |
9 |
2 |
4 |
6 |
8 |

y |
20 |
23 |
15 |
11 |
27 |
21 |
17 |
14 |

1.Calculate and interpret the covariance between x and y.

2.Give a possible reason that the covariance is negative.

3.Calculate the coefficient of correlation, and comment on the relationship between x and y

4.Give a possible reason that the correlation is negative.

A company claims that 10% of the users of a certain sinus drug experience drowsiness. In clinical studies of this sinus drug, 81 of the 900 subjects experienced drowsiness.

- We want to test their claim and find out whether the actual percentage is not 10%. State the appropriate null and hypotheses.
- Is there enough evidence at the 5% significance level to infer that the competitor is correct?
- Construct a 95% confidence interval estimate of the population proportion of the users of this allergy drug who experience drowsiness.
- Explain how to use this confidence interval to test the hypotheses.

Below are monthly rents paid by 30 students who live off campus.

- Find the mean, median, and mode.
- Do the measures of central tendency agree? Explain.
- Calculate the standard deviation.
- Are there outliers or unusual data values?
- Using the Empirical Rule, do you think the data could be from a normal population?

730 |
730 |
730 |
930 |
700 |
570 |

690 |
1,030 |
740 |
620 |
720 |
670 |

560 |
740 |
650 |
660 |
850 |
930 |

600 |
620 |
760 |
690 |
710 |
500 |

730 |
800 |
820 |
840 |
720 |
700 |

Three messenger services deliver to a small town in Oregon. Service A has 60% of all the scheduled deliveries, service B has 30%, and service C has the remaining 10%. Their on-time rates are 80%, 60%, and 40% respectively. Define event O as a service delivers a package on time.

- Calculate P(Aand O).
- Calculate the probability that a package was delivered on time.
- If a package was delivered on time, what is the probability that it was service A?
- If a package was delivered 40 minutes late, what is the probability that it was service B?
- If a package was delivered 40 minutes late, what is the probability that it was service C?

## 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

- 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
- 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.

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