4.4

  1. Point estimate for the average height of active individuals: 171.1
    Median: 170.3

  2. Point estimate for the standard deviation of the heights of active individuals: 9.4 IQR (interquartile range): Q3 - Q1 = 177.8 - 163.8 = 14

  3. Given that the measurements of 180 cm and 155 cm are outside of our IQR, they would be considered unusual (on opposite sides of the spectrum, of course).

  4. The mean and standard deviation of the new sample would most likely not be the same as the ones given above. Given that the samples are random, it’s highly unlikely that the same sample would be selected a second time, so the mean and standard deviation would change.

  5. = 9.4 / sqrt(507) = 0.4174687

9.4/sqrt(507)
## [1] 0.4174687

4.14

  1. False. We are 100% confident that the average spending of these 436 American adults is between 80.31 and 89.11. We are 95% confident that the population mean lies between these values.

  2. I don’t think we have enough information to determine this.

  3. False. Just as means and standard deviations vary per sample, confidence intervals do as well.

  4. True. This is the point I explained in part a.

  5. True. Increasing the confidence level of a confidence interval results in larger intervals. Decreasing the confidence level of a confidence interval results in smaller intervals.

  6. False, this would not be enough to decease the margin of error.

  7. True. The margin of error is 4.4

4.24

  1. All conditions for inference are satisfied. (Independence, sample size, skewness, etc.)

  2. \[ H_o: \mu = 32 \\ H_a: \mu < 32 \]
    \[ z = \frac{30.69 - 32}{4.31}\] = -0.30 pnorm(-0.3) = 0.3820886
    p = 0.382

pnorm(-0.3)
## [1] 0.3820886

  1. Since p(0.382) > 0.10, we cannot reject the null hypothesis. This means we do not have enough evidence to conclude that children first count to 10 successfully when they are 32 months old, on average.

  2. \[SE = \frac{4.31}{\sqrt(36)}\] 90%: 1.645
    30.69 + (1.645 * 0.72) = 30.69 + 1.1844 = 31.8744
    30.69 - (1.645 * 0.72) = 30.69 - 1.1844 = 29.5056
    (29.5056, 31.8744)

  3. The results from our hypothesis test and the confidence interval do agree. We didn’t have enough evidence to conclude that children first count to ten when they are 32 on average, and this number does not fall within our confidence interval.

4.26

  1. \[ H_o: \mu = 100 \\ H_a: \mu \neq 100 \]
    \[ z = \frac{118.2 - 100}{6.5} \\ = 2.8\] pnorm(2.8) = 0.9974449 1 - 0.9974449 = 0.0025551 / 2 = 0.00127755

  2. \[SE = \frac{6.5}{\sqrt(36)} \\ = 1.08333\] 118.2 + (1.282 * 1.083) = 118.2 + 1.388 = 119.588 118.2 - (1.282 * 1.083) = 118.2 - 1.388 = 116.812 (116.812, 119.588)

  3. The results do agree. We rejected the null in favor of the alternative, and the confidence interval reiterates the same message that mothers with gifted children have higher IQ’s than the general population.

4.34

A sampling distribution can be thought of as a relative frequency distribution, displaying a large number of samples drawn from a particular population and their relative frequency to one another.

Shape:
As sample size increases, the sampling distirbution approaches a normal distribution shape.

Center:
As sample size increases, the center is unchanged.

Spread:
As sample size increases, the spread of the sampling distribution will begin to narrow and be less widely distributed.

4.40

  1. \[ z = \frac{10500 - 9000}{1000} \\ = 1.5\] \[pnorm(1.5) = 0.9331928 \\ 1 - 0.9331928 = 0.0668072 \\ p = 0.0668072\]

  2. The distribution of the mean lifespan of 15 light bulbs would be normally distributed, but very widely spread out since the sample size is so small. The center would be around 9,000.

  3. \[SE = \frac{1000}{\sqrt(15)} \\ = 258.1989\] \[ z = \frac{10500 - 9000}{258.1989} \\ = 5.80947\] pnorm(5.80947, 9000, 1000) = 1.189897e-19

Normal <- seq(9000 - (4 * 1000), 9000 + (4 * 1000), length = 15)
Random <- seq(9000 - (4 * 258.1989), 9000 + (4 * 258.1989), length = 15)
Normal2 <- dnorm(Normal, 9000, 1000)
Random2 <- dnorm(Random, 9000, 258.1989)

plot(Normal, Normal2, type="l",col="purple", lty = 2, lwd = 3,
  xlab = "", ylab = "", main="Population vs. Sampling Distributions",
  ylim=c(0,0.0016))
lines(Random, Random2, col="magenta", lwd = 2)

  1. Most likely not, no.

4.48

As sample size increases, our p-value will decrease. So, in this case, when we adjust our sample size from 50 to 500, our p value should be much smaller.