Question

A random sample of 49 measurements from a population with population standard deviation σ₁ = 3 had a sample mean of x₁ = 12. An independent random sample of 64 measurements from a second population with population standard deviation σ₂ = 4 had a sample mean of x₂ = 14. Test the claim that the population means are different. Use level of significance 0.01.

(g) Find a 99% confidence interval for

μ₁ − μ₂.

(Round your answers to two decimal places.)

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REM (rapid eye movement) sleep is sleep during which most dreams occur. Each night a person has both REM and non-REM sleep. However, it is thought that children have more REM sleep than adults†. Assume that REM sleep time is normally distributed for both children and adults. A random sample of n₁ = 10 children (9 years old) showed that they had an average REM sleep time of x₁ = 2.7 hours per night. From previous studies, it is known that σ₁ = 0.5 hour. Another random sample of n₂ = 10 adults showed that they had an average REM sleep time of x₂ = 2.00 hours per night. Previous studies show that σ₂ = 0.7 hour.

What is the value of the sample test statistic? Compute the corresponding z or t value as appropriate. (Test the difference μ₁ − μ₂. Round your answer to two decimal places.)


(iii) Find (or estimate) the P-value. (Round your answer to four decimal places.)

(b) Find a 98% confidence interval for

μ₁ − μ₂.

(Round your answers to two decimal places.)

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Answer 1

1)

Pooled Variance
sp = sqrt(s1^2/n1 + s2^2/n2)
sp = sqrt(9/49 + 16/64)
sp = 0.6585

Given CI level is 0.99, hence α = 1 - 0.99 = 0.01                  
α/2 = 0.01/2 = 0.005, tc = t(α/2, df) = 2.58                  
                  
Margin of Error                  
ME = tc * sp                  
ME = 2.58 * 0.6585                  
ME = 1.699                  
                  
CI = (x1bar - x2bar - tc * sp , x1bar - x2bar + tc * sp)                  
CI = (12 - 14 - 2.58 * 0.6585 , 12 - 14 - 2.58 * 0.6585                  
CI = (-3.70 , -0.30)                  

2)

1)

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: μ1 = μ2
Alternative Hypothesis, Ha: μ1 > μ2

Pooled Variance
sp = sqrt(s1^2/n1 + s2^2/n2)
sp = sqrt(0.25/10 + 0.49/10)
sp = 0.272

Test statistic,
z = (x1bar - x2bar)/sp
z = (2.7 - 2)/0.272
z = 2.57

P-value Approach
P-value = 0.0051

b)

Given CI level is 0.98, hence α = 1 - 0.98 = 0.02                  
α/2 = 0.02/2 = 0.01, tc = t(α/2, df) = 2.33                  
                  
Margin of Error                  
ME = tc * sp                  
ME = 2.33 * 0.272                  
ME = 0.634                  
                  
CI = (x1bar - x2bar - tc * sp , x1bar - x2bar + tc * sp)                  
CI = (2.7 - 2 - 2.33 * 0.272 , 2.7 - 2 - 2.33 * 0.272                  
CI = (0.07 , 1.33)