Assume That The Data Has A Normal Distribution And The Number Of Observations Is Greater Than Fifty. Find The Critical Z Value Used To Test A Null Hypothesis. Alpha = 0.09 For A Right-Tailed Test. (Points : 5) ±1.96 1.34 ±1.34 1.96 2. Find The Value

1. Assume that the data has a normal distribution and the number of observations is greater than fifty. Find the critical z value used to test a null hypothesis.
Alpha = 0.09 for a right-tailed test. (Points : 5)
±1.96
1.34
±1.34
1.96

 

Don't use plagiarized sources. Get Your Custom Essay on
Assume That The Data Has A Normal Distribution And The Number Of Observations Is Greater Than Fifty. Find The Critical Z Value Used To Test A Null Hypothesis. Alpha = 0.09 For A Right-Tailed Test. (Points : 5) ±1.96 1.34 ±1.34 1.96 2. Find The Value
Just from $5/Page
Order Essay

2. Find the value of the test statistic z using z = W4Q3
The claim is that the proportion of drowning deaths of children attributable to beaches is more than 0.25, and the sample statistics include n = 681 drowning deaths of children with 30% of them attributable to beaches. (Points : 5)
3.01
2.85
-2.85
-3.01

3. Use the given information to find the P-value. Also, use a 0.05 significance level and state the conclusion about the null hypothesis (reject the null hypothesis or fail to reject the null hypothesis).
The test statistic in a left-tailed test is z = -1.83. (Points : 5)
0.0672; reject the null hypothesis
0.0336; reject the null hypothesis
0.9664; fail to reject the null hypothesis
0.0672; fail to reject the null hypothesis

 

4. Use the given information to find the P-value. Also, use a 0.05 significance level and state the conclusion about the null hypothesis (reject the null hypothesis or fail to reject the null hypothesis).
With H1: p < 3/5, the test statistic is z = -1.68. (Points : 5)
0.093; fail to reject the null hypothesis
0.0465; fail to reject the null hypothesis
0.0465; reject the null hypothesis
0.9535; fail to reject the null hypothesis

 

5. Formulate the indicated conclusion in nontechnical terms. Be sure to address the original claim.
The owner of a football team claims that the average attendance at games is over 694, and he is therefore justified in moving the team to a city with a larger stadium. Assuming that a hypothesis test of the claim has been conducted and that the conclusion is failure to reject the null hypothesis, state the conclusion in nontechnical terms. (Points : 5)
There is not sufficient evidence to support the claim that the mean attendance is less than 694.
There is sufficient evidence to support the claim that the mean attendance is greater than 694.
There is sufficient evidence to support the claim that the mean attendance is less than 694.
There is not sufficient evidence to support the claim that the mean attendance is greater than 694.

 

6. Assume that a hypothesis test of the given claim will be conducted. Identify the type I or type II error for the test.
A consumer advocacy group claims that the mean mileage for the Carter Motor Company’s new sedan is less than 32 miles per gallon. Identify the type I error for the test. (Points : 5)
Fail to reject the claim that the mean is equal to 32 miles per gallon when it is actually greater than 32 miles per gallon.
Reject the claim that the mean is equal to 32 miles per gallon when it is actually less than 32 miles per gallon.
Reject the claim that the mean is equal to 32 miles per gallon when it is actually 32 miles per gallon.
Fail to reject the claim that the mean is equal to 32 miles per gallon when it is actually less than 32 miles per gallon.

 

7. Find the P-value for the indicated hypothesis test.
In a sample of 88 children selected randomly from one town, it is found that 8 of them suffer from asthma. Find the P-value for a test of the claim that the proportion of all children in the town who suffer from asthma is equal to 11%. (Points : 5)
0.2843
-0.2843
0.2157
0.5686

 

8. Find the P-value for the indicated hypothesis test.
An article in a journal reports that 34% of American fathers take no responsibility for child care. A researcher claims that the figure is higher for fathers in the town of Littleton. A random sample of 225 fathers from Littleton, yielded 97 who did not help with child care. Find the P-value for a test of the researcher’s claim. (Points : 5)
0.0019
0.0015
0.0038
0.0529

9. Find the critical value or values of  CRitVALX2 based on the given information.
H1: sigma > 3.5
n = 14
Alpha= 0.05 (Points : 5)
22.362
5.892
24.736
23.685

10. Find the critical value or values of CritVALX2 based on the given information.
H1: Sigma> 26.1
n = 9
Alpha= 0.01 (Points : 5)
1.646
21.666
20.090
2.088

 

11. Find the number of successes x suggested by the given statement.
A computer manufacturer randomly selects 2850 of its computers for quality assurance and finds that 1.79% of these computers are found to be defective. (Points : 5)
51
56
54
49

 

12. Assume that you plan to use a significance level of alpha = 0.05 to test the claim that p1 = p2, Use the given sample sizes and numbers of successes to find the pooled estimate p-bar Round your answer to the nearest thousandth.
n1 = 570; n2 = 1992
x1 = 143; x2 = 550 (Points : 5)
0.541
0.270
0.520
0.216

 

13. Assume that you plan to use a significance level of alpha = 0.05 to test the claim that p1 = p2. Use the given sample sizes and numbers of successes to find the z test statistic for the hypothesis test.
A report on the nightly news broadcast stated that 10 out of 108 households with pet dogs were burglarized and 20 out of 208 without pet dogs were burglarized. (Points : 5)
z = -0.041
z = -0.102
z = 0.000
z = -0.173

 

14. Solve the problem.

The table shows the number of smokers in a random sample of 500 adults aged 20-24 and the number of smokers in a random sample of 450 adults aged 25-29. Assume that you plan to use a significance level of alpha = 0.10 to test the claim that P1not =P2Find the critical value(s) for this hypothesis test. Do the data provide sufficient evidence that the proportion of smokers in the 20-24 age group is different from the proportion of smokers in the 25-29 age group?
W1T14(Points : 5)
z = ± 1.645; yes
z = ± 1.28; yes
z = ± 1.96; no
z = 1.28; no

 

15. Assume that you plan to use a significance level of alpha = 0.05 to test the claim that p1 = p2. Use the given sample sizes and numbers of successes to find the P-value for the hypothesis test.
n1 = 50; n2 = 75
x1 = 20; x2 = 15 (Points : 5)
0.0032
0.0146
0.1201
0.0001

 

16. Construct the indicated confidence interval for the difference between population proportions p1 – p2. Assume that the samples are independent and that they have been randomly selected.
x1 = 61, n1 = 105 and x2 = 82, n2 = 120; Construct a 98% confidence interval for the difference between population proportions p1 – p2. (Points : 5)
0.456 < p1 – p2 < 0.707
0.432 < p1 – p2 < 0.730
-0.228 < p1 – p2 < 0.707
-0.252 < p1 – p2 < 0.047

 

17. Construct the indicated confidence interval for the difference between the two population means. Assume that the two samples are independent simple random samples selected from normally distributed populations. Do not assume that the population standard deviations are equal.
A researcher was interested in comparing the resting pulse rates of people who exercise regularly and the pulse rates of people who do not exercise regularly. She obtained independent simple random samples of 16 people who do not exercise regularly and 12 people who do exercise regularly. The resting pulse rates (in beats per minute) were recorded and the summary statistics are as follows.
W1T17
Construct a 95% confidence interval for mu1 – mu2, the difference between the mean pulse rate of people who do not exercise regularly and the mean pulse rate of people who exercise regularly. (Points : 5)
-3.22 beats/min < mu1 – mu2 < 11.62 beats/min
-3.55 beats/min < mu1 – mu2 < 11.95 beats/min
-3.74 beats/min < mu1 – mu2 < 12.14 beats/min
-4.12 beats/min < mu1 – mu2 < 14.72 beats/min

 

18. State what the given confidence interval suggests about the two population means.
A researcher was interested in comparing the heights of women in two different countries. Independent simple random samples of 9 women from country A and 9 women from country B yielded the following heights (in inches).
W1T18
The following 90% confidence interval was obtained for mu1 – mu2, the difference between the mean height of women in country A and the mean height of women in country B. -4.34 in. < mu1 – mu2 < -0.03 in
What does the confidence interval suggest about the population means? (Points : 5)
The confidence interval includes only negative values which suggests that the mean height of women from country A is greater than the mean height of women from country B.
The confidence interval includes only negative values which suggests that the two population means might be equal. There doesn’t appear to be a significant difference between the mean height of women from country A and the mean height of women from country B.
The confidence interval includes only negative values which suggests that the mean height of women from country A is smaller than the mean height of women from country B.
The confidence interval includes 0 which suggests that the two population means might be equal. There doesn’t appear to be a significant difference between the mean height of women from country A and the mean height of women from country B.

 

19. Construct the indicated confidence interval for the difference between the two population means. Assume that the two samples are independent simple random samples selected from normally distributed populations. Also assume that the population standard deviations are equal (sigma1 = sigma2), so that the standard error of the difference between means is obtained by pooling the sample variances.
A paint manufacturer wanted to compare the drying times of two different types of paint. Independent simple random samples of 11 cans of type A and 9 cans of type B were selected and applied to similar surfaces. The drying times, in hours, were recorded. The summary statistics are as follows.
W1T19
Construct a 99% confidence interval for mu1 – mu2, the difference between the mean drying time for paint type A and the mean drying time for paint type B. (Points : 5)
-2.73 hrs < mu1 – mu2 < 7.73 hrs
-0.64 hrs < mu1 – mu2 < 5.64 hours
-1.50 hrs < mu1 – mu2 < 6.50 hrs
-2.01 hrs < mu1 – mu2 < 7.01 hrs

 

20. The two data sets are dependent. Find d-Bar to the nearest tenth.
WIT20(Points : 5)
44.1
20.3
33.9
203.4

Order a unique copy of this paper
(550 words)

Approximate price: $22

Basic features
  • Free title page and bibliography
  • Unlimited revisions
  • Plagiarism-free guarantee
  • Money-back guarantee
  • 24/7 support
On-demand options
  • Writer’s samples
  • Part-by-part delivery
  • Overnight delivery
  • Copies of used sources
  • Expert Proofreading
Paper format
  • 275 words per page
  • 12 pt Arial/Times New Roman
  • Double line spacing
  • Any citation style (APA, MLA, Chicago/Turabian, Harvard)

Our guarantees

We are committed to making our customer experience enjoyable and that we are keen on creating conditions where our customers feel secured and respected in their interactions with us.
With our qualified expert team who are available 24/7, we ensure that all our customer needs and concerns are met..

Money Payback-back guarantee

Our refund policy allows you to get your money back when you are eligible for a refund. In such a case, we guarantee that you will be paid back to your credit card. Another alternative we offer you is saving this money with us as a credit. Instead of processing the money back, keeping it with us would be an easier way to pay for next the orders you place

Read more

Zero-plagiarism guarantee

All orders you place on our website are written from scratch. Our expert team ensures that they exercise professionalism, the laid down guidelines and ethical considerations which only allows crediting or acknowledging any information borrowed from scholarly sources by citing. In cases where plagiarism is confirmed, then the costumier to a full refund or a free paper revision depending on the customer’s request..

Read more

Free-revision policy

Quality is all our company is about and we make sure we hire the most qualified writers with outstanding academic qualifications in every field. To receive free revision the Company requires that the Customer provide the request within fourteen (14) days from the first completion date and within a period of thirty (30) days for dissertations.

Read more

Privacy policy

We understand that students are not allowed to seek help on their projects, papers and assignments from online writing services. We therefore strive to uphold the confidentiality that every student is entitled to. We will not share your personal information elsewhere. You are further guaranteed the full rights of originality and ownership for your paper once its finished.

Read more

Fair-cooperation guarantee

By placing an order with us, you agree to the service we provide. We will endear to do all that it takes to deliver a comprehensive paper as per your requirements. We also count on your cooperation to ensure that we deliver on this mandate.

Read more

Calculate the price of your order

550 words
We'll send you the first draft for approval by September 11, 2018 at 10:52 AM
Total price:
$26
The price is based on these factors:
Academic level
Number of pages
Urgency