Question

i need a clear understand how to work out the problems below. Finding thevMean, Standard deviation, and confidence interval

The recorded high temperatures (in degrees Fahrenheit) in June for a sample of 17 US cities are listed below. Assuming the temperatures are normally distributed, find the 92% confidence interval (including units) of the population mean. Give the distribution used and the mean and standard deviation of the sample rounded to the nearest hundredth (two decimal places). Give all digits provided by the calculator for the confidence interval. 70 93 92 78 91 80 101 78 105 102 71 70 86 91 81 84 101 Distribution used: Confidence interval The GPAS for a sample of n 6 freshmen at a certain college are listed below. Assuming the GPAS are normally distributed and that o 0.43, find the 98% confidence interval of the true mean of the freshman class (no units are needed). Give the distribution used and the mean of the sample rounded to the nearest hundredth (two decimal places). Give all digits provided by the calculator for the confidence interval. 3.28 3.07 2.73 2.71 2.86 2.40 Distribution used Confidence interval

Answer 1

Question 1

	Values ( X )	Σ ( Xi- X̅ )2
	78	75.8641
	105	334.5241
	91	18.4041
	80	45.0241
	101	204.2041
	93	39.5641
	92	27.9841
	78	75.8641
	70	279.2241
	91	18.4041
	101	204.2041
	71	246.8041
	81	32.6041
	84	7.3441
	102	233.7841
	70	279.2241
	86	0.5041
Total	1474	2123.5297

Mean X̅ = Σ Xi / n
X̅ = 1474 / 17 = 86.71
Sample Standard deviation S_X = √ ( (Xi - X̅ )² / n - 1 )
S_X = √ ( 2123.5297 / 17 -1 ) = 11.52

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.08 /2, 17- 1 ) = 1.869
86.7059 ± t(0.08/2, 17 -1) * 11.5204/√(17)
Lower Limit = 86.7059 - t(0.08/2, 17 -1) 11.5204/√(17)
Lower Limit = 81.4837
Upper Limit = 86.7059 + t(0.08/2, 17 -1) 11.5204/√(17)
Upper Limit = 91.9281
92% Confidence interval is ( 81.4837 , 91.9281 )

Question 2

	Values ( X )	Σ ( Xi- X̅ )2
	3.28	0.1921
	3.07	0.0521
	2.73	0.0125
	2.71	0.0173
	2.86	0.0003
	2.4	0.1951
Total	17.05	0.4694

Mean X̅ = Σ Xi / n
X̅ = 17.05 / 6 = 2.84

Confidence Interval :-
X̅ ± Z( α /2) σ / √ ( n )
Z(α/2) = Z (0.02 /2) = 2.326
2.8417 ± Z (0.02/2 ) * 0.43/√(6)
Lower Limit = 2.8417 - Z(0.02/2) 0.43/√(6)
Lower Limit = 2.4334
Upper Limit = 2.8417 + Z(0.02/2) 0.43/√(6)
Upper Limit = 3.2500
98% Confidence interval is ( 2.4334 , 3.2500 )