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Dmitry [639]
3 years ago
5

A statistics teacher was reading through a summary of his electric usage for the previous year. Typical high temperatures for ea

ch month were included in the summary and the teacher became curious about the distribution of those temperatures. Here are the data (measured in degrees Fahrenheit).
22, 31, 64, 64, 42, 81, 79, 71, 79, 79, 52, 38


a. Find the median. Interpret this value in context.

b. Calculate the mean high temperature from last year. Show your work.

c. Which measure of center better describes a typical high temperature? Justify your answer.
Mathematics
1 answer:
Fiesta28 [93]3 years ago
3 0

Answer:

64

58.5°

Step-by-step explanation:

Given the data:

22, 31, 64, 64, 42, 81, 79, 71, 79, 79, 52, 38

Rearranged data: 22, 31, 38, 42, 52, 64, 64, 71, 79, 79, 79, 81

The median temperature will be :

1/2(n + 1)th term.

n = 12 (number of samples)

0.5 * (12 + 1)th

= 0.5 (13th)

= 6.5 th term

(6th + 7th term) / 2 = (64 + 64) / 2 = 64

The middle temperature value is 80°F

The mean temperature value :

ΣX /n

= sum(22, 31, 38, 42, 52, 64, 64, 71, 79, 79, 79, 81) / 12

= 702 / 12

= 58.5° F

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A recent national survey found that high school students watched an average (mean) of 7.2 movies per month with a population sta
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Answer:

1. Option A

2. Option D) Reject null hypothesis if z < –1.645

3. z = -4.90

4. P-value = 0.00001

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 7.2

Sample mean, \bar{x} = 6.5

Sample size, n = 49

Alpha, α = 0.05

Population standard deviation, σ = 1.0

a) First, we design the null and the alternate hypothesis

H_{0}: \mu \geq 7.2\text{ movies per month}\\H_A: \mu < 7.2\text{ movies per month}

Option A) is the correct answer

We use one-tailed(left) z test to perform this hypothesis.

c) Formula:

z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }

Putting all the values, we have

z_{stat} = \displaystyle\frac{6.5 - 7.2}{\frac{1}{\sqrt{49}} } = -4.90

b) Now, z_{critical} \text{ at 0.05 level of significance } = -1.645

Rejection rule:

If calculated z-statistic is less than the critical z-value, we reject the null hypothesis.

Option D) Reject null hypothesis if z < –1.645

d) Now, we calculate the p-value with the help of standard normal table.

P-value = 0.00001

Since the p-value is lower than the significance level, we fail to accept the null hypothesis and reject it.We accept the alternate hypothesis.

We conclude that the college students watch fewer movies a month than high school students.

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3 years ago
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