Question

Review Questions from Previous Fu- 1. Jessie thinks that per eople gene- 226 Bonus Review 3 - Chapters 9-12 For each of the following problems (except the questions that simply ask you to fill in the tables), conduct the appropriate hypothesis test or construct the appropriate confidence interval. In addition, for each word problem, conclude with a sentence that explains what the results MEAN. 1a. Tanya knows that fiber is important for digestive health. Therefore, she thinks that people who take fiber pills daily will have fewer stomach aches. To test this, Tanya asks 64 participants to take a fiber pill each day for one month and to record how many stomach aches they suffer in the following month. Tanya knows that people who don't take fiber pills daily generally suffer about 2 stomach aches per month, on average. If Tanya's sample suffers 1.8 stomach aches per month, on average, with a standard deviation of 2.1, can Tanya conclude that taking fiber pills daily will result in significantly fewer stomach aches? 16. Construct the two-tailed 90% Confidence interval that estimates the true population average number of stomach aches suffered in one month by people who take fiber pills daily. Make sure to include a concluding statement explaining the meaning of what you calculated. table

Answer 1

1a)

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: μ = 2
Alternative Hypothesis, Ha: μ < 2

Rejection Region
This is left tailed test, for α = 0.1 and df = 63
Critical value of t is -1.295.
Hence reject H0 if t < -1.295

Test statistic,
t = (xbar - mu)/(s/sqrt(n))
t = (1.8 - 2)/(2.1/sqrt(64))
t = -0.762

P-value Approach
P-value = 0.2245
As P-value >= 0.1, fail to reject null hypothesis.

b)

sample mean, xbar = 1.8
sample standard deviation, s = 2.1
sample size, n = 64
degrees of freedom, df = n - 1 = 63

Given CI level is 90%, hence α = 1 - 0.9 = 0.1
α/2 = 0.1/2 = 0.05, tc = t(α/2, df) = 1.669

ME = tc * s/sqrt(n)
ME = 1.669 * 2.1/sqrt(64)
ME = 0.438

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (1.8 - 1.669 * 2.1/sqrt(64) , 1.8 + 1.669 * 2.1/sqrt(64))
CI = (1.36 , 2.24)
Therefore, based on the data provided, the 90% confidence interval for the population mean is 1.36 < μ < 2.24 which indicates that we are 90% confident that the true population mean μ is contained by the interval (1.36 , 2.24)