HW_C4_606

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 themedian is 170.3.

What is the point estimate for the standard deviation of the heights of active individuals? What about the IQR?
```
Standard deviation is 9.4, IQR is 163.8 - 177.8.
```

Is a person who is 1m 80cm (180 cm) 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 has a lot sample fall between 175cm to 180cm. 155cm may be unsually short, since it is on the very left tail of the hist.

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.
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Not the same but should be similar
```
(e)The sample means obtained are point estimates for the mean height of all active individuals, if the sample of individuals is equivalent to a simple random sample. What measure do we use to quantify the variability of such an estimate (Hint: recall that SDx¯=??n???)?
```
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. the 95% confident interval refer to the whole univese not to the sample mean. ```
This confidence interval is not valid since the distribution of spending in the sample is right skewed.
```
True, since it is not normal distribution.
```
95% of random samples have a sample mean between $80.31 and $89.11.
```
False. 95% refers to the whole populaiton.
```
We are 95% confident that the average spending of all American adults is between $80.31 and $89.11.
```
True.
```
A 90% confidence interval would be narrower than the 95% confidence interval since we don’t need to be as sure about our estimate.
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True, it includes less sample size.
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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.
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Fales, need to be 9 time larger, from the se function.
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The margin of error is 4.4.
```
True
```

### 4.24 Gifted children, Part I
(a) Are conditions for inference satisfied?

Yes, it is more than 30, and is random sample

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  = 32
HA < 32
alpha = 0.1
SE = sd / sqrt(n) = 0.72
Z= (30.69-32)/0.72 = -1.82

Interpret the p-value in context of the hypothesis test and the data.

p(z<-1.82)= 0.0341 , Since the p-value is less than the significance level, we can reject the null hypothesis

Calculate the 90% confidence interval for the average age at which gifted children first count to 10 successfully.

lower = mean-sigma*sd
upper = mean+sigma*sd
mean=30.69
sd=4.31
Confidence Interval is 29.50845 to 31.87155.

Do your results from the hyposthesis test and the confidence interval agree? Explain.
```
Yes, the confiedence interval does not contain 32 months, so we rejected the null.
```

4.26 Gifted Children Part 2

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=100 Ha!=100
the p-value is less than the significance level, we can reject the null hypothesis

z <- (118.2-100)/(6.5/6)
p <- 2*pnorm(-abs(z))
p

## [1] 2.44044e-63

Calculate the 90% confidence interval for the average IQ of mothers of gifted children.
```
lower = 118.2 - 1.645*(6.5/6) =116.4179 
upper = 118.2 + 1.645*(6.5/6) =119.9821
```
Do your results from the hypothesis test and the confidence interval agree? Explain.
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Yes, because 100 is not in the confidence interval
```

4.34 CLT

Define the term “sampling distribution” of the mean, and describe how the shape, center, and spread of the sampling distribution of the mean change as sample size increases.

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

4.40 CFLBs

What is the probability that a randomly chosen light bulb lasts more than 10,500 hours?

p =(1 - pnorm(10500, mean = 9000, sd = 1000))=0.0668072

Describe the distribution of the mean lifespan of 15 light bulbs.
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It is nomral.
```
What is the probability that the mean lifespan of 15 randomly chosen light bulbs is more than 10,500 hours?
```
Z =(10500-9000)/(1000/15^0.5)
1 - pnorm(Z) =3.133452e-09
```
Sketch the two distributions (population and sampling) on the same scale.
Could you estimate the probabilities from parts (a) and (c) if the lifespans of light bulbs had a skewed distribution?
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No
```

4.48 Same observation, different Sample Size.

The larger the Z-score, the smaller the p-value