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inessss [21]
3 years ago
7

The average hourly wage for a high school student in Illinois is $8.25 per hour with a standard deviation of $.25. Assume hourly

wages vary normally.
A) Determine the margin of error for this scenario.
B) For the previous question you just answered, construct a 95% confidence interval for the hourly wage for a high school student.
C) For the 95% confidence interval you just constructed, does an hourly wage of $9.15 fall within that acceptable range?
Mathematics
1 answer:
serious [3.7K]3 years ago
7 0

Answer:

Step-by-step explanation:

Hello!

The variable of interest is

X: hourly wage of a high school student

X~N(μ;σ²)

X[bar]= $8.25

σ= $0.25

For a 95% CI for the average hourly wage for<u> 1 </u>high school student:

For this exercise you have to consider the sample size n=1

a) The margin of error of the CI is calculated as:

d= Z_{1-\alpha /2} * σ

1-α: 0.95 ⇒ α:0.05

1-α/2:0.975

Z_{1-\alpha /2}= Z_{0.975}= 1.96

d= 1.96 * 0.25= 0.49

b)

The formula for the interval is

X[bar] ± Z_{1-\alpha /2} * σ

[8.25 ± 0.49]

[7.76; 8.74]

With a 95% you'd expect the interval $[7.76; 8.74] to include the average hourly wage of one high school student in Illinois.

c)

Te calculated interval is $ [7.76; 8.74] the value $9.15 does nor fall within that acceptable range.

I hope this helps!

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Complete question:

The manager of a supermarket would like to determine the amount of time that customers wait in a check-out line. He randomly selects 45 customers and records the amount of time from the moment they stand in the back of a line until the moment the cashier scans their first item. He calculates the mean and standard deviation of this sample to be barx = 4.2 minutes and s = 2.0 minutes. If appropriate, find a 90% confidence interval for the true mean time (in minutes) that customers at this supermarket wait in a check-out line

Answer:

(3.699, 4.701)

Step-by-step explanation:

Given:

Sample size, n = 45

Sample mean, x' = 4.2

Standard deviation \sigma = 2.0

Required:

Find a 90% CI for true mean time

First find standard error using the formula:

S.E = \frac{\sigma}{\sqrt{n}}

= \frac{2}{\sqrt{45}}

= \frac{2}{6.7082}

SE = 0.298

Standard error = 0.298

Degrees of freedom, df = n - 1 = 45 - 1 = 44

To find t at 90% CI,df = 44:

Level of Significance α= 100% - 90% = 10% = 0.10

t_\alpha_/_2_, _d_f = t_0_._0_5_, _d_f_=_4_4 = 1.6802

Find margin of error using the formula:

M.E = S.E * t

M.E = 0.298 * 1.6802

M.E = 0.500938 ≈ 0.5009

Margin of error = 0.5009

Thus, 90% CI = sample mean ± Margin of error

Lower limit = 4.2 - 0.5009 = 3.699

Upper limit = 4.2 + 0.5009 = 4.7009 ≈ 4.701

Confidence Interval = (3.699, 4.701)

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