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3 years ago

Jack is placing glass bottles into boxes to recycle. He can fit 12 bottles in each box.

1 answer:

6 0

His estimation is not reasonable because if you can only fit 12 bottles in a box and he has 10 boxes then he could only fit 120 out of the 200 bottles so he is short on boxes

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A car ruled 160 meters in 3 hours how far did it travel in 5 hours

Bess [88]

5.3333333 repeating for one hour and multiply that by 5 and get........... 26.5 I think It's to dark for me to right anything I'm going off the top of my head

5 0

3 years ago

Exercise 6.22 provides data on sleep deprivation rates of Californians and Oregonians. The proportion of California residents wh

kherson [118]

Answer:

a. The alternative hypothesis H₀: p'₁ ≠ p'₂ is accepted

b. Type I error

Step-by-step explanation:

Proportion of California residents who reported insufficient rest = 8.0%

Proportion of Oregon residents who reported insufficient rest = 8.8%

p'₁ = 0.08 * 11545 =923.6

p'₂ = 0.088 * 4691=412.81

σ₁ = $\sqrt{n*p_1*q_1} = \sqrt{n*p_1*(1-p_1)}$ = $\sqrt{11545*0.08*(1-0.08)}$ = 29.15

σ₂ = $\sqrt{n*p_2*q_2} = \sqrt{n*p_2*(1-p_2)}$ = $\sqrt{4691*0.088*(1-0.088)}$ = 19.40

Samples size of California residents n₁ = 11,545

Samples size of Oregon residents n₂ = 4,691

Hypothesis can be constructed thus

Let our null hypothesis be H ₀: p'₁ = p'₂

and alternative hypothesis H ₐ: p'₁ ≠ p'₂

Then we have

$z =\frac{(p'_1 -p'_2)-(\mu_1-\mu_2)}{\sqrt{\frac{\sigma^2_1}{n_1}+\frac{\sigma^2_2}{n_2} } }$

The test statistics can be computed by

t₀ = $\sqrt{\frac{n_1n_2(n_1+n_2-2)}{n_1+n_2} } *\frac{p_1'-p_2'}{\sqrt{(n_1-1)\sigma_1^2+(n_2-1)\sigma_2^2} }$ = 1104.83

c from tables is P(T ≤ c) = 1 - α where α = 5% and c = 1.65

since t₀ ≥ c then then the hypothesis is rejected which means the alternative hypothesis is rejected

b. Type I error, rejecting a true hypothesis

5 0

3 years ago

Assume that adults have IQ scores that are normally distributed with a mean of mu equals 105 and a standard deviation sigma equa

Sindrei [870]

Answer:

0.7698

Step-by-step explanation:

If you call your random variable $X$ , then what you are looking for is

$P(87 \leq X \leq 123)$

because you want the probability of $X$ being <em>between 87 and 123.</em>

We need a table with of the normal distribution. But we can only find the table with $\mu = 0$ and $\sigma = 1$ . Because of that, first we need to <em>normalize </em>our random variable:

$Z = \frac{X - \mu}{\sigma} = \frac{X - 105}{15}$

(you can always normalize your variable following the same formula!)

now we can do something similar to our limits, to get a better expression:

$\frac{87 - 105}{15} = \frac{-18}{15} = -1.2$

$\frac{123 - 105}{15} = \frac{18}{15} = 1.2$

And we transform our problem to a simpler one:

$P(87 \leq X \leq 123) = P(-1.2 \leq Z \leq 1.2) = P(Z \leq 1.2) - P(Z \leq -1.2)$

(see Figure 1)

From our table we can see that $P(Z \leq 1.2) = 0.8849$ (this is represented in figure 2).

Remember that the whole area below the curve is exactly 1. So we can conclude that $P(Z \geq 1.2) = 0.1151$ (because 0.8849 + 0.1151 = 1). We also know the normal distribution is symmetric, then