For the following​ information, determine whether a normal sampling distribution can be​ used, where p is the population​ proportion, α is the level of​ significance, p is the sample​ proportion, and n is the sample size. If it can be​ used, test the claim.   ​Claim: p≥0.34​; α=0.08. Sample​ statistics: p=0.30​, n=150       Question content area bottom Part 1 Let q=1−p and let q=1−p. A normal sampling distribution ▼     be used​ here, since ▼     ▼   greater than or equals≥ less than< 5 and ▼   nqnq n ModifyingAbove q with caretnq ▼   less than< greater than or equals≥ 5.

Q: Suppose that you are given the plot below that shows the sampling distribution of Y, sample mean…

A: As the sample size increases the sampling distribution closer to the normal distribution. Also, the…

Q: etermine whether a normal sampling distribution can be used for the following sample statistics. If…

A: For the given data Perform z test for two proportion

Q: Determine the sample size necessary to estimate p for the following information. a. E = 0.04, p is…

A: Given Hello. Since your question has multiple parts, we will solve first 3sub parts question for…

Q: Decide whether the normal sampling distribution can be used. If it can be used, test the claim about…

A: The population proportion is 0.21. The sign of the alternative hypothesis is not equal to, so this…

Q: For the following​ information, determine whether a normal sampling distribution can be​ used, where…

A: Given  p&gt;0.29​(claim) α=0.06 Sample​ statistics p=0.36​ n=425

Q: Given the following parameters for a sampling distribution of sample proportions, calculate the…

A:

Q: The average height of women in the freshman clas at UA has historically been 5'4" with a standard…

A: Consider a random variable X, that defines the height of women in the freshman class at the UA.

Q: Find the standard error of the mean for each sampling situation assuming a normal population o = 54,…

A: From the provided information, Population standard deviation (σ) = 54 Sample size (n) = 9

Q: Decide whether the normal sampling distribution can be used. If it can be used, test the claim about…

A: Given : p^ = 0.21   and   n =100 Here, np= 21 and  n (1-P)= 79  both are greater than 5. Thus we can…

Q: 10.1 10.6 9.8 9.9 2 9.7 9.5 10.3 9.9 3 10.1 10.7 9.2 10 4 9.9

A: sample number 1 2 3 4 1 10.1 10.6 9.8 9.9 2 9.7 9.5 10.3 9.9 3 10.1 10.6 9.2 10 4 9.9 9.8…

Q: How large a sample needs to be selected if we wish our population proportion estimate to be within…

A:

Q: Arandom sample of n= 100 observations is selected from a population with u = 29 and o = 23. Complete…

A: As the student wants answers for c,d,e &amp; f only. So, providing solution of those only.

Q: 2. For a normal distribution, find the z-score location that divides the distribution as follows:…

A:

Q: For a sample with s = 9, a score of X = 67 corresponds to z = +1.00. What is the sample mean?

A: Given that: Sample standard deviation, s=9 Score, X=67 Z-score, z=1 To find: Value of sample mean

Q: For which sample size (n) and sample proportion (6) can a normal curve be used to approximate the…

A: Answer :- Find  for which sample size (n)  and sample proportion (P^)  can a normal…

Q: For the sample difference scores below, determine if the sample differs from D = 0. Use α = 0.01.…

A: To testH0:D=0H1:D≠0

Q: or r=0.7789 and sample size=10, the corresponding t value is:"     2.512     3.512     3.034…

A: r = 0.7789 , n = 10

Q: Decide whether the normal sampling distribution can be used. If it can be used, test the claim about…

A: Given,p⏞=0.19n=200α=0.05H0: p=0.22H1:  p≠0.22

Q: Find the mean, variance, and standard deviation of the sampling distribution of the sample…

A: Given: n = 241 p = 0.6 Formula Used: Mean = np Variance = np(1-p) Standard deviation = np(1-p)

Q: onsider a population proportion p = 0.32. [You may find it useful to reference the z table.]   a.…

A: Solution: Given information: p = 0.32 Population proportion  n= 13  sample size  n= 70 Sample size…

Q: estimate of for the following. E = .17, ? = .90, confidence level = 99%

A: (1) We have given that,  Population proportion (p) = 0.90, margin of error (E) = 0.17 and…

Q: primary school students. Teach- 7.6 In 1985, the state of Tennessee carried out a st ers and…

A: Hello! As you have posted more than 3 sub parts, we are answering the first 3 sub-parts.  In case…

Q: a. Determine the sampling distribution of the sample mean for samples of size 3. Interpret your…

A: Hey there! Thank you for posting the question. Since your question has more than 3 parts, we are…

Q: Find the values of the UCL and LCL for the (x-bar) chart for this data, and round your answers to 3…

A: Refer to the table of sample values in Section 8.2 Exercise 6 of the text.Find the values of the UCL…

Q: If we collect a large sample of blood platelet counts and if our sample includes a single outlier,…

A: If we collect a large sample of blood platelet counts and if our sample includes a single…

Q: Determine whether a normal sampling distribution can be used for the following sample statistics. If…

A: points to pass for normal approximation:1) experiment consists of a sequence of n identical trials2)…

Q: Calculate the standardized test statistic. The standardized test statistic is (Round to two decimal…

A: The summary of statistics is, x¯=1308.11,σ=196.31,n=200,μ=1280

Q: Decide whether the normal sampling distribution can be used. If it can be used, test the claim about…

A: The single proportion z-test statistic is, Z=p^-p0p0 (1-p0)nwhere p^=sample proportionp0=claimed…

Q: Use technology to help you test the claim about the population mean, p, at the given level of…

A: Given that the standardized test statistic is 2.56. Here, the test is right tail test.

Q: Decide whether the normal sampling distribution can be used. If it can be used, test the claim about…

A: The random variable X follows binomial distribution. We have to use normal approximation to binomial…

Q: Suppose 65% of all college students find watching TV a waste of time. The population proportion is…

A: Consider that p is proportion of all college students find watching TV a waste of time.

Q: H Find z when = The z-score for the sampling distribution is given by z = 15, 4 = 15, o, = 22, %3D…

A: As givev,  x¯ = 15 μx =15 σx=22 n = 16 It is required to obtain the z-score.

Q: 3. For each of the possible samples, determine the following (provide a solution only for 1 sample):…

A: Given The sample is given below: 8,10,4

Q: The z-score for the sampling distribution is given by z = Find z when z = 15, Hz = 15, oz = 22, %3D…

A: Given that, X¯ = 15μX = 15σX = 22n = 16

Q: When determining sample size for a proportion, we need to know p. If we have no prior information…

A: When determining sample size for a proportion, we need to know p.

Q: Decide whether the normal sampling distribution can be used. If it can be used, test the claim about…

A: We have to test claim.

Q: For which sample size (n) and population parameter (p) can a normal curve be used to approximate the…

A: Given that  If np&gt;10 and nq&gt;10 then it follows normal

Q: Please only do 4 and 5 of the additinal questions portion

A: The question is about hypo. testing Given : Sample mean facial WHR ( x¯ ) = 1.96 Sample std.…

Q: For the following​ information, determine whether a normal sampling distribution can be​ used, where…

A: Givenclaim is p&gt;0.29α=0.06p^=0.36n=425

Q: Decide whether the normal sampling distribution can be used. If it can be used, test the claim about…

A: Given Alpha=0.10 X1=108 n1=128 X1=32 n1=195

Q: Use technology to help you test the claim about the population mean, p, at the given level of…

A: The summary of statistics is, x¯=1308.11,σ=196.31,n=200,μ=1280 The null and alternative hypothesis…

Q: For the data set given, what is the standard error of the sampling distribution of mean difference…

A: Consider, Subject 1st 2nd Difference 1 10 11 -1 2 4 6 -2 3 7 9 -2 4 6 5 1…

Q: Use technology to help you test the claim about the population mean, u, at the given level of…

A: Given,   Sample size = 300 Sample mean = 1267 Population standard deviation = 208.13   Under the…