## Question

###### STUN. Herwinyes 8C3/8C4 Elementary Statistics laps Elvira Jinadu & I 07/30/20 11:31 PM Homework: Section 6.2...

STUN. Herwinyes 8C3/8C4 Elementary Statistics laps Elvira Jinadu & I 07/30/20 11:31 PM Homework: Section 6.2 Homework y Save stics Score: 0 of 1 pt 8 of 7 (5 complete) HW Score: 17.14%, 1.2 of 7 pts 6.2.25-T Question Help 987 The state test scores for 12 randomly selected high school seniors are shown on the right. Complete parts (a) through (c) below. Assume the population is normally distributed 1429 693 1222 723 741 1446 728 838 549 949 626 (a) Find the sample mean. x = 910.9 (Round to one decimal place as needed.) (b) Find the sample standard deviation S = 304.7. (Round to one decimal place as needed.) (c) Construct a 90% confidence interval for the population mean A 90% confidence interval for the population mean is (0) (Round to one decimal place as needed.) Enter your answer in the edit fields and then click Check Answer All parts showing Clear All Check Answer

Homework: Section 6.2 Homework tar You Save Score: 0 of 1 pt 7 of 7 (6 complete) HW Score: 20.71%, 1.45 of 7 p 6.2.26-T Question Help 3.3 The grade point averages (GPA) for 12 randomly selected college students are shown on the right. Complete parts (a) through (c) 2.5 below. 1.6 Assume the population is normally distributed. 2.5 0.3 0.6 1.2 2.3 2.70 4.0 3.7 3.2 (a) Find the sample mean x 2.33 (Round to two decimal places as needed.) (b) Find the sample standard deviation s- 1.19 (Round to two decimal places as needed.) (c) Construct a 10% confidence interval for the population mean . A 90% confidence interval for the population mean is ( (Round to two decimal places as needed.) Enter your answer in the edit fields and then click Check Answer, All parts showing Clear All Check Answer

## Answers

solution:

6.2.25-T(a)

(b)

(c)

---------------------------------------------------

6.2.26-T

(a)(b)

(c)

The sample mean X is computed as follows: i ==> X = 10931 12 910.917 i=1Also, the sample variance s is o==+(*-:(x)) - (197876 Mogna = 92849.538 Therefore, the sample sandartd deviation s is s= V s2 = 92849.538 = 304.7129,8+ the mome) The provided sample mean is Ħ = 910.917 and the sample standard deviation is s = 304.712. The size of the sample is n = 12 and the required confidence level is 90%. The number of degrees of freedom are df = 12 - 1=11, and the significance level is a = 0.1. Based on the provided information, the critical t-value for a = 0.1 and df = 11 degrees of freedom is te = 1.796. The 90% confidence for the population mean is computed using the following expression CI = x te xs Vn Therefore, based on the information provided, the 90 % confidence for the population mean u is CI = ( 910.917 1.796 x 304.712 -,910.917+ 12 1.796 x 304.712 12 (910.917 - 157.971,910.917 + 157.971) = (752.946, 1068.888)The sample mean X is computed as follows: 27.9 12 2.325 1=1Also, the sample variance sº is $? +($*-:(**)) - 2 (1995-1990) -- = 1.408 n Therefore, the sample sandartd deviation s is s=32 = V1.408 = 1.186The provided sample mean is X = 2.325 and the sample standard deviation is s = 1.186. The size of the sample is n = 12 and the required confidence level is 99%. The number of degrees of freedom are df = 12 - 1=11, and the significance level is a = 0.01. Based on the provided information, the critical t-value for a = 0.01 and df = 11 degrees of freedom is te = 3.106 The 99% confidence for the population mean pe is computed using the following expression te x 8 CI = x te XS (* X + vn Therefore, based on the information provided the 99 % confidence for the population mean u is 3.106 x 1.186 3.106 x 1.186 CI = (2.325 ,2.325+ V12 V12 = (2.325 - 1.063, 2.325 +1.063) = (1.262, 3.388)

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