Question

According to the Bureau of Labor Statistics, the mean salary for registered nurses in Kentucky was $58,605. The distribution of salaries is assumed to be normally distributed with a standard deviation of $5,688. Someone would like to determine if registered nurses in Ohio have a greater average pay. To investigate this claim, a sample of 297 registered nurses is selected from the Ohio Board of Nursing, and each is asked their annual salary. The mean salary for this sample of 297 nurses is found to be $58,709.232.

- Completely describe the sampling distribution of the sample
mean salary when samples of size 297 are selected.
- mean: μ¯yμy¯ = ____
- standard deviation: σ¯yσy¯ = ___ (round your answer to 4 decimal places)
- shape: the distribution of ¯y is ____ (not normally distributed normally distributed) because ____. (the sample size is not large the population of salaries is normally distributed the population of salaries is not normally distributed the sample size is large)

- What conjecture has been made?
- The mean salary for registered nurses in Ohio is $58,709.232.
- The mean salary for registered nurses in Ohio is greater than $58,605.
- The mean salary for registered nurses in Ohio is $58,605.
- The mean salary for registered nurses in Ohio is greater than $58,709.232.

- Using the distribution described in part a, what is the
probability of observing a sample mean of 58,709.232 or more?
- z =___ (round to 2 decimal places)
- probability =___ (include 4 decimal places)

- Based on the probability found, what conclusion can be reached?
- The probability would be classified as ____ (large small) . So, there ____ (is is not) evidence to support the conjecture that the mean salary for registered nurses in Ohio is greater than $58,605.

In 2011, the number of text messages sent and received by teenage girls (ages 12 – 18) was strongly right skewed. The mean number of messages sent and received each day was 165 with a standard deviation of 45 messages. Suppose we assume teenage boys (ages 12 – 18) send and receive the same number of messages daily.

- Completely describe the sampling distribution of the sample
mean number of text messages sent and received when samples of 259
teenage boys are selected.
- mean: μy¯ = ____
- standard deviation: σ¯y= ____ (round your answer to 4 decimal places)
- shape: the distribution of ¯y is ______ (normally distributed not normally distributed) because ______ . (he sample size is large the sample size is not large the population of number of text messages sent is normally distributed the population of number of text messages sent is not normally distributed)

- Using the distribution described in part a, what is the
probability of observing a sample mean of 163.852 or less?
- z = ____(round to 2 decimal places)
- probability = ____ (include 4 decimal places)

- Classify the probability found in part b using the rule of
thumb discussed in class. SELECT ONE
- The probability would be classified as small.
- The probability would be classified as large.

- Based on the probability found, what conclusion can be reached?
- There ____ (is is not) sufficient evidence to conclude the mean number of text messages sent by male teenagers is _____ (greater than less than) 165.

Black Friday - the annual shopping tradition the day after Thanksgiving - is often the day which puts retailers "in the black." According to a CNN Money report, consumers spent an average of $360.44 on Black Friday in 2010 with a standard deviation of $236.70.

- Draw and label a normal curve which would be used to describe
the Black Friday expenditures. Based on the values calculated,
would it be reasonable to assume the money spent is normally
distributed? SELECT ONE
- It is not reasonable to assume the amount of money spent by Black Friday shoppers is normally distributed.
- It is reasonable to assume the amount of money spent by Black Friday shoppers is normally distributed

- Completely describe the sampling distribution of the sample
mean Black Friday expenditure when samples of size 64 are selected.
- Mean: μ¯y = ___
- Standard deviation: σ¯y= ___ (round to 4 decimal places)
- Shape:the distribution of ¯y is _____(normally distributed not normally distributed) because ____ (the population of expenditures is normally distributed the population of expenditures is not normally distributed the sample size is not large the sample size is large)

- Using the distribution described in part b, what is the
probability of observing a sample mean of $441.101 or more?
- z = ____ (round to 2 decimal places)
- probability = ____ (include 4 decimal places)

- Based on the probability found, what conclusion can be reached?
- The probability would be classified as ___ (small large) . So, there ___ (is is not) sufficient evidence to conclude the mean amount spent by customers on Black Friday is greater than 360.44.

Answer #1

1)

mean: μ¯y = $58,605

standard deviation: σ¯y = σ / sqrt(n) = 5,688 / sqrt(297) = 330.0510

shape: the distribution of ¯y is normally distributed because he sample size is large the sample size is not large the population of number of text messages sent is normally distributed

d)

Probability > 0.05, So The probability would be classified as small. So, there is evidence to support the conjecture that the mean salary for registered nurses in Ohio is greater than $58,605.

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