A hypothesis test procedure uses a simple randomsample to compare an unknown population mean, μ,to 75,000. A simple random sample of size 23 yields asample mean of x =76,677. The underlyingpopulation is known to be normally distributed with apopulation standard deviation of 12500. Compute theP-value for the following hypothesis test using three-decimal place accuracy.Ho : μ = 75000HA: μ >75000

GIvensample size(n)=23Mean(x)=76677standard deviation(σ)=12500The null and alternative hypothesis…

Answered: A hypothesis test procedure uses a…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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A sample of size n = 36 is drawn from a population whose standard deviation is σ = 15.6. The sample mean is x = 94.5. Construct a 95% confidence interval.
Consider a population with a mean of 150 and a standard deviation of s = 27. A random sample of 36 was taken from this population. Then the standard deviation of X-bar is:
200 people were randomly sampled and asked what they regularly eat for breakfast or lunch. Each person was identified as either a consumer or a non consumer of high-fiber cereals, and the number of calories consumed at lunch was measured and recorded. These data are summarized below; Consumer of high fiber cereals Non consumer of high fiber cereals η1 =41 η2 = 159 Mean 1 =603 Mean 2 =639 Stanadard deviation 1 = 110 Standard deviation 2 = 141 If the scientist claims that people who eat high fiber cereals for breakfast do consume on average fewer calories for lunch than people who don’t eat high fiber cereals for breakfast, and if it is true, high fiber cereal manufacturer will be able to claim another advantage of eating their products-potential weight reduction for dieter. REQUIRED Are there sufficient evidence at 5% significance level to support the above claim?
The null and alternate hypotheses are: He: P₁ =H₂ H₁ H₁ #P₂ A random sample of 15 observations from the first population revealed a sample mean of 350 and a sample standard deviation of 12. A random sample of 17 observations from the second population revealed a sample mean of 342 and a sample standard deviation of 15. The population variances are assumed to be equal. At the 0.10 significance level, is there a difference in the population means? a. Is this a one-tailed or a two-tailed test? O One-tailed test. OTwo-tailed test. b. State the decision rule. (Negative amounts should be indicated by a minus sign. Round your answers to 3 decimal places.) The decision rule is to reject H0 if t
Given a normally distributed data set whose mean is 25 standard deviation is 5, what values of x would a z-score of 1.2 have?
A population has a mean μ = 73 and a standard deviation o = 28. Find the mean and standard deviation of a sampling distribution of sample means with sample size n = 257. Hx ox (Simplify your answer.) (Type an integer or decimal rounded to three decimal places as needed.)
pls. answer using 7 steps:-Parameter of interest-Null hypothesisAlternative hypothesis -Test statistic -Reject H0 if: -Computations: -Conclusion
Two independent samples have been selected, 7777 observations from population 1 and 5858 observations from population 2. The sample means have been calculated to be ?⎯⎯⎯1=13.4x¯1=13.4 and ?⎯⎯⎯2=11.3x¯2=11.3. From previous experience with these populations, it is known that the variances are ?21=27σ12=27 and ?22=29σ22=29. For the hypothesis test of ?0:(?1−?2)=1.8H0:(μ1−μ2)=1.8 and ??:(?1−?2)≠1.8Ha:(μ1−μ2)≠1.8 Use ?=0.04α=0.04. (a) Compute the test statistic. ?=z= (b) Find the approximate p-value?−?????=p−value= The final conclustion is A. There is not sufficient evidence to reject the null hypothesis that (?1−?2)=1.8(μ1−μ2)=1.8. B. We can reject the null hypothesis that (?1−?2)=1.8(μ1−μ2)=1.8 and accept that (?1−?2)≠1.8(μ1−μ2)≠1.8.
In a random sample of size 36 we observe a sample mean of 25 and a variance of 144. If we test the hypotheses below with a significance level of 0.05. Ho: µ = 20 HẠ :4 > 20 what will be the calculated test statistic?
A random sample of n = 25 is obtained from a population with variance ?2, and the sample mean is computed to be ?̅= 70. Consider the null hypothesis ?0: ? =80 versus the alternative hypothesis ?1: ? < 80. Compute the p-value when the population variance is ?2 = 600.
Use z scores to compare the given values.Based on sample data , newborn males have weights with a mean of 3242.5 g and a standard deviation of 687.2 g . Newborn females have weights with a mean of 3085.4 and a standard deviation of 829.1 g . Who has the weight that is more extreme relative to the group from which they came : a male who weighs 1500 g or a female who weighs 1500 ?
The average dues paid by students at state universities across the country were $2236 per student last year. The average for a random sample of 500 students this year is $2350. Given the following hypotheses, Ho: 4 = 2236 (The average fees are the same as last year.) Ha: M + 2236 (The average fees are not the same as last year.) compute the p-value for the sample result. Assume the population standard deviation is $724. If the significance level is 5%, can the null hypothesis be rejected?

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