Week 4 Homework

4.6 Art After School

  1. Sampling Distribution

  2. I would expect the shape of this distribution to be symmetric. I would expect this distribution to be symmetric because it seems like the number of eggs laid would be a normal curve.

  3. In order to calculate the variability or standard deviation of this distribution we must use the normal approximation of the binomial distribution.

n=100 p=.93

mean = np = 93 SD = sqrt(np(1-p)) = sqrt(93*.07) = 2.55

d)The variablity of the new distribtion would likely increase, because there is a larger number of people being surveyed there is more variability in their answers.

4.12 Mental Health

  1. We are 95% confidence that the true population mean for the number of days in the past 30 days that an individual’s mental health was not good falls between 3.4 and 4.24 days.

  2. 95% confidence means that 95% of the sample means for the population parameters will fall between 3.4 and 4.24 days.

  3. A 99% confidence interval would be larger becuase 99% of the sample means must fall within the interval, rather than 95%

  4. If the same survey were done with 500 Americans the standard deviation would be lower because there would be less variability in the answers.

4.18 Identify Hypotheses

  1. Ho: Calorie intake after calorie counts started = Calorie intake before calorie counts started

Ha: Calorie intake after calorie counts started < Calorie intake before calorie counts started

  1. Ho: Average GRE verbal reasoning score has not changed Ha: Average GRE verbal reasoning score has changed

4.24 Gifted Children

  1. Yes. Sample>30 and symmetric

Ho: Average age >= 32

Ha:Average age < 32

First calculate Z:

z = (30.69-32)/(4.31/sqrt(36))

Then the p:

pnorm(-1.823666,0,1)
## [1] 0.03410129

Since the p-value .034 is less than the alpha of .10 we reject the null hypothesis.

  1. The p-value suggests that there is enough evidence to infer that gifted children can count to 10 at an earlier age.

  2. 90% Confidence

30.69-1.64*4.31/sqrt(36)
## [1] 29.51193
30.69+1.64*4.31/sqrt(36)
## [1] 31.86807

The 90% CI is 29.51, 31.87.

  1. Yes. Becuase 32 is not within the confidence interval, we can see that it is not part of the population mean.

4.30 Testing for Food Safety

  1. Ho: Restaurant met regulations

Ha: Restaurant did not meet regulations

  1. A type 1 error in this context would mean the inspector decides they do not meet regulations, even though they do.

  2. A type 2 error would mean the inspector decides they do meet regulatoins, even though they do not.

  3. Type 1 error is more problematic for the owner because he is punished for no reason.

  4. Type 2 error would be worse for diners becuase their food is not up to regulation standards even though they are told it is.

  5. Strong evidence. Becuase any sign of health concerns could be bad for the consumer of the product.

4.32 True or False

  1. False.

  2. True

  3. False

  4. False

4.38 Identify Distributions

  1. B. This is the most skewed, with a lot of variability due to the fact that it is one random sample with 500 observations.

  2. C. This is more normal, with very little variability due to its large sample but small number of observations.

  3. A. This is the most normal and has the least variability because it is a large sample with a lot of observations.

4.44 Nearsighted

mean = p = 0.08,

SE = sqrt(0.08∗0.92) / 194 = 0.0195

SD= sqrt(15.52(1-.08)) = 3.77

Ho: Nearsightedness in children = 8%

Ha: Nearsightedness in children (not)= 8%

  1. 21/194 or 10.8%

  2. z = (mean-x)/(sd/sqrt(n))

(.108 - .08)/ (3.77/sqrt(194))
## [1] 0.1034469
pnorm(0.1034469,0,1)
## [1] 0.5411959

Since the p-value is greater than the alpha we fail to reject the null hypothesis that nearsightedness in children = 8%.