HW4

4.4 Heighrs of adults

What is the point estimate for the average height of active individuals? What about the median?

The point estimate for the average height of active individual is 171.1. and the point estimate for the median is 170.3.

What is the point estimate for the standard deviation of the heights of active individuals? What about the IQR?
```
The point estimate for the standard deviation is 9.4. 
```

Q3 <- 177.8
Q1 <- 163.8
IQR <- Q3 - Q1
IQR

## [1] 14

Is a person who is 1m 80cm (180cm) tall considered unusually tall? And is a person who is 1m 55cm (155cm) considered unusually short? Explain your reasoning.
```
180cm is not unusually tall since it is only 1 s.d away the mean. 155 could be unusually short since it is 2 s.d away from the mean.
```
The researchers take another random sample of physically active individuals. Would you expect the mean and the standard deviation of this new sample to be the ones given above? Explain your reasoning.
```
The mean and the standard deviation could be different from above but it should be similar data.
```
The variability is measured by the standard error.
```
SE = 9.4 / sqrt(507) = 0.4175
```

4.14 Thanksgiving Spending

We are 95% confident that the average spending of these 436 American adults is between $80.31 and $89.11.

False. We are 95% confident that the average spending of the total Amaerican adults "population", not "sample of 436"has spent between $80.31 and $89.11

This confidence interval is not valid since the distribution of spending in the sample is right skewed.
```
False. The sample data is slightly right skewed but mostly distributed normal.
```

95% of random samples have a sample mean between $80.31 and $89.11.

False. It is 95% confidence interval not the percentage of the sample.

We are 95% confident that the average spending of all American adults is between $80.31 and $89.11.
```
True. As explained for (d)
```
A 90% confidence interval would be narrower than the 95% confidence interval since we don’t need to be as sure about our estimate.
```
True. the lower the CI, the narrower the area covered in the plot.
```
In order to decrease the margin of error of a 95% confidence interval to a third of what it is now, we would need to use a sample 3 times larger.
```
False. If we use 3 times the sample, the SE = 0.054. Which is still higher than the third of what is now (0.031)  
```
The margin of error is 4.4
```
True
```

4.24 Gifted children, Part I

Are conditions for inference satisfied?

Since the sample size is 36 which is more than 30, it would be inference statistics.

Suppose you read online that children first count to 10 successfully when they are 32 months old, on average. Perform a hypothesis test to evaluate if these data provide convincing evidence that the average age at which gifted children first count to 10 successfully is less than the general average of 32 months. Use a significance level of 0.10.
```
H0 : mean = 32
HA: mean < 32
alpha = 0.1
SE = sd / sqrt(n)= 4.31 / 6 = 0.72
Z= (30.69-32)/0.72 = -1.82
p(z<-1.82)= 0.0341 
```
Interpret the p-value in context of the hypothesis test and the data.
```
p(z<-1.82)= 0.0341  which is less than 0.1, we reject H0, accept Ha.
```
Calculate the 90% confidence interval for the average age at which gifted children first count to 10 successfully.
```
lower = 30.69 -(1.28*0.72)=29.77
upper = 30.69+(1.28*0.72)= 31.61
CI = (29.77, 31.61)
```

Do your results from the hyposthesis test and the confidence interval agree? Explain.

The results agree because the range of the confidence interval does not contain 32 months.

4.26 Gifted Children

Perform a hypothesis test to evaluate if these data provide a convincing evidence that the average IQ of mothers of gifted children is different that the average IQ for the population at large, which is 100. Use a significance level of 0.10.
```
H0: mean = 100
Ha: mean > 100
alpha = 0.1
SE = sd / sqrt(n) = 6.5 / 6 = 1.0833
z = (118.2 -100)/ 1.0833 = 16.80
since Z is so high, even without p-value test, we still can reject H0, accept Ha.
```

Calculate the 90% confidence interval for the average IQ of mothers of gifted children.

lower= 100 - (1.2816 * 1.0833) = 116.8116
upper = 100+(1.2816 * 1.0833) = 119.5884
CI = (116.8116, 119.5884)

Do your results from the hypothesis test and the confidence interval agree? Explain.

The results agree because 100 is not in the confidence interval, and we rejected the null hypothesis.

4.34 CLT

Sampling distribution of the mean is the distribution of mean values calculated from repeated sampling of a population. As the sample size increases, the shape of the distribution tends to be more normal, the spread gets smaller.

4.40 CFLBs

What is the probability that a randomly chosen light bulb lasts more than 10,500 hours?
```
Z = (10500-9000)/1000 = 1.5
P(z > 1.5)= 1-0.933=6.68%
```
Describe the distribution of the mean lifespan of 15 light bulbs.
```
The distribution is approximately normal.
```
What is the probability that the mean lifespan of 15 randomly chosen light bulbs is more than 10,500 hours?
```
SE = 1000/sqrt(15)=258.20
Z = (10500-9000)/ 258.2 = 5.8
P(Z>5.8)=0.0003
```
Sketch the two distributions (population and sampling) on the same scale.

library(tidyverse)
mu <- 9000
s <- 1000
se <- s / sqrt(15)
x <- seq(mu - (4 * s), mu + (4 * s), length=100)
hx <- dnorm(x, mean=mu, sd=s)
df <- data.frame(name="Population", x, hx)
smpl <- seq(mu - (4 * se), mu + (4 * se), length=100)
hxSmpl <- dnorm(smpl, mean=mu, sd=se)
df <- rbind(df, data.frame(name="Sampling", x=smpl, hx=hxSmpl))
ggplot() +
    geom_line(data=df, aes(x=x, y=hx, color=name)) +
    labs(title="Distribution of CFLBs")

Could you estimate the probabilities from parts (a) and (c) if the lifespans of light bulbs had a skewed distribution?
```
No. 
```

4.48 Same observation, different Sample Size.

When sample size increases, SE will be decrease, and Z score will be increase, P will decrease.

HW4

Qixing Li

March 2, 2019