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For each αand observed significance level (p-value) pair, indicate whether the null hypothesis would be rejected.

a.α=0.5,p-value=.10

b.α=0.10,p-value=.05

c.α=0.01,p-value=.001

d.α=0.25,p-value=.05

e.α=0.10,p-value=.45

Short Answer

Expert verified

The null hypothesis would be rejected for every case except the last one because the given p-values are greater than the value in the last one.

Step by step solution

01

General rule for each case

P-value stands for probability value. It indicates how likely it is that a result occurred by chance alone.

Here, αis the level of significance.

If the p-value is less than the level of significance, reject the null hypothesis. Otherwise do not reject the null hypothesis.

02

Finding whether the null hypothesis will be rejected or not for part a.

a. Given that,α=0.5

The p-value is 0.10

Here, the p-value is less than the αvalue.

Therefore, we reject the null hypothesis

03

Finding whether the null hypothesis will be rejected or not for part b.

b. Given that, α=0.10

The p-value is 0.05

Here, the p-value is less than the αvalue.

Therefore, we reject the null hypothesis.

04

Finding whether the null hypothesis will be rejected or not for part c.

c. Given that,α=0.01

The p-value is 0.001

Here, the p-value is less than the αvalue.

Therefore, we reject the null hypothesis.

05

Finding whether the null hypothesis will be rejected or not for part d.

d. Given that,α=0.25

The p-value is 0.05

Here, the p-value is less than the αvalue.

Therefore, we reject the null hypothesis.

06

Finding whether the null hypothesis will be rejected or not for part e.

e. Given that, α=0.10

The p-value is 0.45

Here, the p-value is greater than the αvalue.

Therefore, we do not reject the null hypothesis.

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Most popular questions from this chapter

Studies have established that rudeness in the workplace can lead to retaliatory and counterproductive behaviour. However, there has been little research on how rude behaviours influence a victim’s task performance. Such a study was conducted, and the results were published in the Academy of Management Journal (Oct. 2007). College students enrolled in a management course were randomly assigned to two experimental conditions: rudeness condition (students) and control group (students). Each student was asked to write down as many uses for a brick as possible in minutes. For those students in the rudeness condition, the facilitator displayed rudeness by generally berating students for being irresponsible and unprofessional (due to a late-arriving confederate). No comments were made about the late-arriving confederate to students in the control group. The number of different uses for brick was recorded for each student and is shown below. Conduct a statistical analysis (at α=0.01) to determine if the true mean performance level for students in the rudeness condition is lower than the actual mean performance level for students in the control group.

The data is given below

Control Group:

124516217201920191023160491317130212117311119912185213015421211101311361013161228191230


Rudeness Condition:

411181196511912757311191110789107114135478381591610071513921310

Question: Forecasting daily admission of a water park. To determine whether extra personnel are needed for the day, the owners of a water adventure park would like to find a model that would allow them to predict the day’s attendance each morning before opening based on the day of the week and weather conditions. The model is of the form

where,

y = Daily admission

x1 = 1 if weekend

0 otherwise

X2 = 1 if sunny

0 if overcast

X3 = predicted daily high temperature (°F)

These data were recorded for a random sample of 30 days, and a regression model was fitted to the data.

The least squares analysis produced the following results:

with

  1. Interpret the estimated model coefficients.
  2. Is there sufficient evidence to conclude that this model is useful for predicting daily attendance? Use α = .05.
  3. Is there sufficient evidence to conclude that the mean attendance increases on weekends? Use α = .10.
  4. Use the model to predict the attendance on a sunny weekday with a predicted high temperature of 95°F.
  5. Suppose the 90% prediction interval for part d is (645, 1,245). Interpret this interval.

Find the following probabilities for the standard normal random variable z:

a.P(0<z<2.25)b.P(-2.25<z<0)b.P(-2.25<z<1.25)d.P(-2.50<z<1.50)e.P(z<-2.33orz>2.33)

Corporate sustainability of CPA firms. Refer to the Business and Society (March 2011) study on the sustainability behaviors of CPA corporations, Exercise 2.23 (p. 83). Recall that the level of support for corporate sustainability (measured on a quantitative scale ranging from 0 to 160 points) was obtained for each of 992 senior managers at CPA firms. The accompanying Minitab printout gives the mean and standard deviation for the level of support variable. It can be shown that level of support is approximately normally distributed.

a. Find the probability that the level of support for corporate sustainability of a randomly selected senior manager is less than 40 points.

b. Find the probability that the level of support for corporate sustainability of a randomly selected senior manager is between 40 and 120 points.

c. Find the probability that the level of support for corporate sustainability of a randomly selected senior manager is greater than 120 points.

d. One-fourth of the 992 senior managers indicated a level of support for corporate sustainability below what value?

Descriptive Statistics: Support

Variables

N

Mean

StDev

Variance

Minimum

Maximum

Range

Support

992

67.755

26.871

722.036

0.000

155.000

155.000

Assume that x is a binomial random variable with n = 1000 andp = 0.50. Use a normal approximation to find each of the following probabilities:

a. P(x>500)

b.P(490x<500)

c.P(x>550)

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