Question

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1. The World Health Organization estimates that 5% of all adults in sub-Saharan Africa are living with HIV/AIDS. A survey takes a random sample of 1600 adults from all over sub-Saharan Africa and finds that 72 have HIV/AIDS.

a) What is the sample proportion and what is the population proportion?

b) What is the mean of the sampling distribution?

c) What is the standard deviation of the sampling distribution?

d) What is the probability that a random sample of 1600 adults in sub-Saharan Africa would have less than 4.5% living with HIV/AIDS?

2. In a random sample of 36 students at a college, the average IQ was x = 107. Assume that the IQ of a student in college follows a Normal distribution with unknown mean and standard deviation σ = 16.

a) Give a 90% confidence interval for μ.

b)Give a 99% confidence interval for µ.

c) Which confidence interval has a larger margin of error? Why does this make sense?

3) A researcher collects a random sample of size n from a population with standard deviation σ and, from the data collected, computes a 95% confidence interval for the mean of the population. Which of the following would produce a new confidence interval with a smaller width (smaller margin of error) based on these same data?
A. Increase σ.
B. Use a lower confidence level.
C. Use a smaller sample size.

Explain why you chose your answer.

B) The IQ scores for adults in the entire population have an approximately normal distribution with mean 100 and standard deviation 15. The researcher’s question in “Do college students aged 20 to 25 years old have a higher mean IQ than the rest of the population?” What are the appropriate null and alternative hypothesis a researcher would have to use if they chose to do an experiment to try to answer this question? Explain.

C) A survey on the number of hours spent working out was given to professors from the US. Assume the standard deviation of the number of hours spent by all professors in the US was found to be 1.94 hours. At a 98% confidence level, how many professors would have to be surveyed in order to bring the margin of error down to 0.1?

Answer 1

1) (a) The population proportion, p = 5% = 0.05

The sample proportion = = 72/1600 = 0.045

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(b) The Mean () is given by n * p = 1600 * 0.05 = 80

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(c) The Standard deviation () is given by Sqrt(n * p * (1-p)) = SQRT(1600 * 0.05 * 0.95) = 8.7178

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(d) 4.5% of 1600 = 72, Therefore to find the Probability, P(X < 72)

The Z score = (X - )/ = (72 - 80)/8.7178 = -0.92

The required probability (from the normal distribution tables) = 0.1788

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2) Given = 107, = 16, n = 36

Since is known and n is > 30, we use Z critical values.

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(a)The 95% CI: Zcritical at = 0.05 is 1.96.

The Confidence interval is given by ME

The Lower Limit = 107 - 5.23 = 101.77

The Upper Limit = 107 + 5.23 = 112.23

The 95% Confidence Interval is (101.77 , 112.23)

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(a)The 99% CI: Zcritical at = 0.01 is 2.576.

The Confidence interval is given by ME

The Lower Limit = 107 - 6.87 = 100.13

The Upper Limit = 107 + 6.87 = 113.87

The 99% Confidence Interval is (100.13 , 113.87)

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(c) As we see, the 99% CI has a larger Margin of Error (ME = 6.87), versus the ME of the 95% CI (ME = 5.23).

This is because when we keeping all other factors unchanged (i.e and n), the only factor which is changing is the Zcritical value, which is in the numerator of the ME.

This means the ME will increase or decrease as the Zcritical changes. As the Confidence level increases, the Zcritical also increases (and vice versa is also true), and therefore the ME also increases.

Here since we changed from a 95% CI to a 99% CI, the Zcritical increased and therefore ME also increased.

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3) (A) The width of the data depends on The Margin of error (ME), which is given by

Therefore the ME is dependent on the following.

(i) Zcritical, which is in the numerator. As Zcritical increases or decreases, ME will also increase or decrease respectively. Z critical depends on Confidence level. When confidence level increases (or decreases), the Zcritical also increases (or decreases) respectively.

(ii) The population standard deviation (), is also in the numerator. Hence if is increased or decreased, the ME will also increase or decrease respectively.

(iii) Sample Size (n). This is in the denominator, and hence when n increases, ME decreases and vice versa i.e when n decreases, ME will increase.

Therefore for a smaller width, we should Use a lower condfidence level. (Option B)

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(B) When creating questions on Hypothesis, we have the null and the Alternative. The Null will always have the equal to (=) sign. The alternative is always trying to prove that the Null is not true, and therefore it comes as either one tailed tests with a greater than (>) ir a lesser than (<) sign. We also have a 2 tailed alternative which comes with a not equal to () sign.

Also the null and the alternative will test if the population means are the same or have they changed, based on the sample that we are using.

Here the researcher is trying to test if college students aged 20 to 25 years old have a higher mean IQ than the rest of the population. Therefore the Hypothesis:

H0: = 100 : College students of 20 to 25 years of age have a mean IQ equal to 100.

Ha: > 100 : College students of 20 to 25 years of age have a mean IQ greater than 100.

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(C) SAMPLE SIZE FOR MEANS

Given ME = 0.1, = 1.94 hours, = 0.02

The Zcritical (2 tail) at = 0.02 is 2.326

The ME is given by :

Squaring both sides we get: (ME)² = (Z critical)² * ²/n

Therefore n = (Zcritical * /ME)² = (2.326*1.94/0.1)² = 2036.21

Therefore n = 2037 (Taking it to the next whole number)

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