2. State University requires an admission test of all its students. A simple ran

Question

2. State University requires an admission test of all its students. A simple randorm sample of 121 students from the University was selected to estimate the average te score of all University students. The average (mean) for the sample was found tob 23.4 with a sample standard deviation of 3.65. (4 points each) a) Construct a 95% confidence interval for the average (mean) test score of all University students. Be sure to tell which kind of interval you have chosen What does the 95% confidence interval that you calculated in (a) tell you about the avera (mean) test score of all University students? (b) Which of the following would result in a narrower confidence interval for the average (mean) test score of all University students? (select.ALL that apply) (i) Maintain the confidence level at 95%, but increase the sample size to 151. (ii) Maintain the confidence level at 95%, but decrease the sample size to 91 (iii) Keep the sample size at 121, but increase the confidence level to 99%. (iv) Keep the sample size at 121, but decrease the confidence level to 90%. (c) when we say "95% confidence" in a 95% confidence interval, we believe that (select, one) (d), 95% ofthe intervals constructed using this process based on samples (ofthe san (i) (i) There is a 95% probability that a particular confidence interval will include the (iii) 95% of the possible population means will be included in the interval. (iv) size) from this population will include the population mean. sample mean. 95% of the possible sample means will be included in the interval.

Explanation / Answer

Solution:

(a)We have to find 95% confidence interval for population mean.

We are given,

Sample mean = M = 23.4

Sample standard deviation = s = 3.65

Sample size = n = 121

Level of significance = a = 0.05

Now we have to find Z at a = 0.05

Using normal probability table,

Z = 1.96

Formula for CI is,

= M ± Z(sM)

Calculation

M = 23.4
t = 1.96
sM = (3.652/121) = 0.33

= M ± Z(sM)
= 23.4 ± 1.96*0.33
= 23.4 ± 0.65

95% CI is [22.75, 24.05]

(b) There is 95% of the possible population means will be included in the interval [22.75, 24.05]

(c) Option (i) is correct.

(d) Option ( i i i ) is correct.

Done

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