4.4

a) Mean: 171.1 Median: 170.3

b) SD: 9.4 IQR: 7.5

c)

Mean - 2 x SD = 152.3 Mean + 2 x SD = 189.9

The 180 CM person is well within 2 SD of the mean and barely past the IQR. They are not unusually tall. The 155 CM person is still within 2 SD of the mean, but well past the IQR and based on the histogram would fall in a low percentile. However, they still would not be considered unusually short by most measures, since they would be above the 2nd percentile.

d)

The sample size from above is fairly large and it does appear to follow a normal distribution, so it’s not skewed by outliers. Therefore we can be farily confident that the mean and SD of the next sample will be close to this one’s. We can therefore make inferences about how close we could expect the next sample to be

e)

It would be the standard error (SE) = 0.4174687

4.14

**a)* False - We are 100% sure on the mean of these adults.

b) False - After 30 sample the sum of lognormals doesn’t quite converge to normal, but after 400, it should

c) False - nothing is said about individual sample from the above facts

d) True - This sample should be representative of the population

e) True - This will represent a smaller area under the normal curve

f) False - We would need a sample 9 times larger

g) True - That’s one way of looking at the confidence interval. The mean +- 4.4

4.24

a) Independence appears to be satisfied, as it is a large city and the children were likely taken from different schools. The distribution exhibits some skew, so I’m not entirely sure a sample of 36 is adequate. If budgetary restrictions allowed for it, I would suggest more samples be taken.

b) The two must differ by at least 1.28 SE to pass a one tailed test at 10%

Null: The average age of counting is not less for gifted children than that of the population.
Alternative: The average age of counting is lower for gifted children

(32 - 30.69)/(4.31 x 6) = 1.9489559

This meets the criteria for a signficant difference

c)

P value: 0.0256503

This means that if the average age of counting for the whole population and the sample both belonged to the same normal distribution, there would be a 2.6% chance of observing a mean from the sample of gifted children less than or equal to 30.6

d)

29.5083417 to 31.8716583

e) Yes, 32 falls outside of the confidence interval. It would be possible for them to agree and have it be at the edge of the invteral, since it was a one tailed test and only needed to differ by 1.28 SE

4.26

a) Null hypothesis: The mean of the mother’s IQ does not differ from the population mean Alternative hypothesis: The mean of the mother’s IQ does differ from the population

(118.2 - 100)/(6.5/6) = 16.8

This meets the requirement of differing from the mean by at least 1.645 SE needed for a two tailed 10% test

b) 116.4179167 to 119.9820833

c) These results agree, as 100 is well out of the confidence interval

4.34

The sampling distribution of the mean is the distribution of the mean of samples from the population. If we take random samples from the population and calculate their mean, that distribution will get more normal, the more samples are taken. The easiest example is sampling from a uniform using two samples. It’s unlikely we’ll observe two extreme observations, so the tails will have less density. If we imagine discrete uniforms for convenience, it’s like the sum of two dice. There are more possible ways to get a total of 7, which is the mean. As you add more dice, the distriubtion will look more and more normal.

4.40

a) 0.0668072

b) This will be nearly normal with mean 9000 and SD = 1000/sqrt(15) = 258.1988897

c) 1500/(1000/sqrt(15)) = 5.809475 Probability: 3.133452210^{-9}

d)

norm1 <- rnorm(100000, mean = 9000, sd = 1000)

norm2 <- rnorm (100000, mean = 9000, sd = 1000/sqrt(15))

norms <- data.frame(sample = norm1, means = norm2)

ggplot(norms) + geom_density(aes(x = sample)) + geom_density(aes(x = means))

e) Not part A, but possibly part C depending on the skew. Given survival time for non living things is usually Weibull then yes.

4.48

Your p value should decrease. If we assume you plugged 50 into the formula to get the standard error for the sample, changing it to 500 will decrease it by root 10. You will then see the test value differ from the sample by more standard errors, making the p value lower.