f it has a different average sore. lat nline test and assume the standard deviat

Question

f it has a different average sore. lat nline test and assume the standard deviation is -12.0 d diern dingnostic test at a college had a mean of 72.6 points. A study is conde about an online diagnostic test to determine the (unk points score of the o (a) In a random sample of 36 average was confidence interval for at 95% confilmenlevel, using one- test-takers, the sample average was 68.5 points. Obtain a two-sided level, using one-decimal place of rounding. (b) Interpret the confidence interval in the contoxt of the problem. Moreover, decide if we can concltade that the mean score has changod from the old mean of 72.6 points. (c) Determine the minimum number of students to be sampled so that the corresponding margin of error at 99% confidence!evel will be at most 3 points. 3. A study is conducted in a city to estimate the proportion, p, of the bomewners who their houses during the next four years. A random sample of 54 homeowners is selected. (a) If 12 of the 54 homeowners (in the sample) planned to sell their houses during the next four years, construct a two-sided confidence interval for p at 90% level of confidence.

Explanation / Answer

Question 2

Part a

We are given

Xbar = 68.5

= 12

n = 36

c = 95%

Critical Z value = 1.96 (by using z-table)

Confidence interval = Xbar -/+ Z*/sqrt(n)

Confidence interval = 68.5 -/+ 1.96*12/sqrt(36)

Confidence interval = 68.5 -/+ 1.96*2

Confidence interval = 68.5 -/+ 3.92

Lower limit = 68.5 – 3.92 = 64.6

Upper limit = 68.5 + 3.92 = 72.4

Confidence interval = (64.6, 72.4)

Part b

We are 95% confident that the mean score of traditional algebra diagnostic test is lies between 64.6 and 72.4.

The value for old mean 72.6 is not included in the above confidence interval, so there is sufficient evidence to conclude that mean score has changed from the old mean of 72.6 points.

Part c

We are given

E = 3

= 12

c = 99%

Critical Z value = 2.5758

Sample size = n = (Z*/E)^2 = (2.5758*12/3)^2 = 106.1559

Required sample size = 107

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