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RideAnS [48]
2 years ago
13

Suppose that your statistics professor returned your first midterm exam with only a z-score written on it. She also told you tha

t a histogram of the scores was mound shaped and approximately symmetric. How would you interpret each of the following z-scores?
a. 2.2
b. 0.4
c. 1.8
d. 1.0
e. 0
Mathematics
1 answer:
lilavasa [31]2 years ago
5 0

Answer:

A <em>z</em>-score specifies the number of standard deviations an observation is from the mean.  

Step-by-step explanation:

A z-score (aka, a standard score) specifies the number of standard deviations an observation is from the mean.  

The formula to compute the z-score is, Z = \frac{X - \mu}{\sigma}, where <em>X</em> = value of the random variable, <em>µ</em> = mean, <em>σ</em> = standard deviation.

The random variable <em>X</em> follows a Normal distribution with parameters <em>µ</em> and <em>σ</em>².

(a)

A <em>z</em>-score of 2.2 implies that the score in the first midterm exam is 2.2 standard deviations above the mean.

(b)

A <em>z</em>-score of 0.4 implies that the score in the first midterm exam is 0.4  standard deviations above the mean.

(c)

A <em>z</em>-score of 1.8 implies that the score in the first midterm exam is 1.8  standard deviations above the mean.

(d)

A <em>z</em>-score of 1.0 implies that the score in the first midterm exam is 1.0  standard deviations above the mean.

(e)

A <em>z</em>-score of 0 implies that the score in the first midterm exam is 0  standard deviations above the mean. That is the score in the first midterm exam is same as the mean score.

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Abe,a plumber, charges $50 per hour plus $75 for making an in-home visit. The Plumbing Service charges $65 per hour but no set f
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Steps:

let “x” be number of hrs
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2 years ago
The U.S. Bureau of Labor Statistics reports that 11.3% of U.S. workers belong to unions (BLS website, January 2014). Suppose a s
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Answer:

a) Null hypothesis:p \leq 0.113  

Alternative hypothesis:p > 0.113  

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

Replacing we got:

z=\frac{0.13 -0.113}{\sqrt{\frac{0.113(1-0.113)}{400}}}=1.074  

The p value for this case would be given by:

p_v =P(z>1.074)=0.141  

c) For this case we see that the p value is higher than the significance level of 0.05 so then we have enough evidence to FAIL to reject the null hypothesis and we can't conclude that the true proportion workers belonged to unions is significantly higher than 11.3%

Step-by-step explanation:

Information given

n=400 represent the random sample taken

X=52 represent the  workers belonged to unions

\hat p=\frac{52}{400}=0.13 estimated proportion of workers belonged to unions

p_o=0.113 is the value that we want to test

\alpha=0.05 represent the significance level

Confidence=95% or 0.95

z would represent the statistic

p_v represent the p value

Part a

We want to test if the true proportion of interest is higher than 0.113 so then the system of hypothesis are.:  

Null hypothesis:p \leq 0.113  

Alternative hypothesis:p > 0.113  

Part b

The statistic is given by:

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

Replacing we got:

z=\frac{0.13 -0.113}{\sqrt{\frac{0.113(1-0.113)}{400}}}=1.074  

The p value for this case would be given by:

p_v =P(z>1.074)=0.141  

Part c

For this case we see that the p value is higher than the significance level of 0.05 so then we have enough evidence to FAIL to reject the null hypothesis and we can't conclude that the true proportion workers belonged to unions is significantly higher than 11.3%

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