Use the standard normal distribution or the t distribution to construct a 90% co

Question

Use the standard normal distribution or the t distribution to construct a 90% confidence interval for the population mean. Justify your decison f ne er s ibu tion can be used, explain why. Interpret the results n a random sample of 44 people, the mean body mass index (BMI) was 26.1 and the standard deviation was 6.18 y A. Use a t-distribution because the sample is random, n 2 30, and is unknown. O B. Use a normal distribution because the sample is random, the population is normal, and is known. C. Use a t-distribution because the sample is random, the population is normal, and is unknown D. Use a normal distribution because the sample is random, n 230, and is known. O E. Neither a normal distribution nor a t-distribution can be used because either the sample is not random, or n

Explanation / Answer

Given :-

Mean (M) = 26.1

Standard deviation (s) = 6.18

Sample size (n) = 44

90% confidence level at alpha= 10% and df = n-1 = 43

then critical value is t = 1.68

Therefore the 90% confidence interval for population mean is,

mu = M ± t * (s /n )

mu = 26.1 ± 1.68*( 6.18 / 44)

mu = 26.1 ± 1.57

=> 26.1 - 1.57 < mu < 26.1 + 1.57

24.53 < mu < 27.67

90% confidence interval is ( 24.53 , 27.67 ).

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