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Ad libitum [116K]

2 years ago

8

The monthly utility bills in a city at e normally distributed, with a mean $100 and a standard deviation of $13. find the probab

ility that a randomly selected utility bill is (a) a less than $70, (b) between $85 and $120, and(c) more than $140.

1 answer:

dmitriy555 [2]2 years ago

5 0

Answer:

how's your day

Step-by-step explanation:

how's your day

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Step-by-step explanation:

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3 years ago

The mayor of a town has proposed a plan for the annexation of an adjoining community. A political study took a sample of 900 vot

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Answer:

$z=\frac{0.75 -0.72}{\sqrt{\frac{0.72(1-0.72)}{900}}}=2.00$

Now we can calculate the p value. Since is a bilateral test the p value would be:

$p_v= P(Z>2) =0.0228$

Since the p value is lower than the significance level of 0.05 we have enough evidence to conclude that the true proportion of residents favored annexation is higher than 0.72 or 72%

Step-by-step explanation:

Information given

n=900 represent the random sample selected

$\hat p=0.75$ estimated proportion of residents favored annexation

$p_o=0.72$ is the value that we want to test

represent the significance level

z would represent the statistic

$p_v$ represent the p value

Hypothesis to test

The political strategist wants to test the claim that the percentage of residents who favor annexation is above 72%.:

Null hypothesis: $p\leq 0.72$

Alternative hypothesis: $p > 0.72$

The statistic for this case is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

Replacing the data given we got:

$z=\frac{0.75 -0.72}{\sqrt{\frac{0.72(1-0.72)}{900}}}=2.00$

Now we can calculate the p value. Since is a bilateral test the p value would be:

$p_v= P(Z>2) =0.0228$

Since the p value is lower than the significance level of 0.05 we have enough evidence to conclude that the true proportion of residents favored annexation is higher than 0.72 or 72%

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