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- Rosie is an aging sheepdog in Montana who gets regular checkups from her owner, the local veterinarian. Let x be a random variable that represents Rosie’s resting heart rate (in beats per minute). From past experience, the vet knows that x has a normal distribution with σ= 12. The vet checked the manual and found that for dogs of this breed, μ= 115beats per minute. Over the past six weeks(n=6), Rosie’s sample mean hear rate was x̄=105.0. The vet is concerned that Rosie’s heart rate is slowing and is less than 115.0. Does the data indicate that this is the case? Use a level of significance α= .05 to conduct a hypothesis test. You must include the following: i) null hypothesis ii) alternate hypothesis iii) test statistic (z or t value –remember the information in the problem will help determine if you use a z or t) iv) P-Value compared to the level of significance(α). v) Conclusion (Do you reject or fail to reject the vet’s claim?)Let x represent the number of mountain climbers killed each,year. The long-term variance of x is approximately o Suppose that for the past 12 years, the variance has been s = 118.3. = 136.2 Use a 1% level of significance to test the claim that the recent variance for number of mountain-climber deaths is less than 136.2 Find a 90% confidence interval for the population variance. (b) Find the value of the chi-square statistic for the sample. (Round your answer to two decimal places.) What are the degrees of freedom?Let x be a random variable that represents the weights in kilograms (kg) of healthy adult female deer (does) in December in a national park. Then x has a distribution that is approximately normal with mean ? = 62.0 kg and standard deviation ? = 9.0 kg. Suppose a doe that weighs less than 53 kg is considered undernourished. (c) To estimate the health of the December doe population, park rangers use the rule that the average weight of n = 70 does should be more than 59 kg. If the average weight is less than 59 kg, it is thought that the entire population of does might be undernourished. What is the probability that the average weight x for a random sample of 70 does is less than 59 kg (assuming a healthy population)? (Round your answer to four decimal places.) Suppose park rangers captured, weighed, and released 70 does in December, and the average weight was x = 63.8 kg. Do you think the doe population is undernourished or not? Explain. A. Since the sample average is above the mean, it…
- Suppose X is a non-standard normal variable, with a mean of 139, and std deviation of 15 (i.e. X~N(139,15)). You have an observation (z) on a standard normal scale (i.e. Z~N(0,1)). If z = 1.29, what is it on an X scale? How can I calculate this using excel & by hand?Suppose X is a random variable of a normal distribution with E(X)=2 and Var(X)=4. Find P(X<4)Please do a, e, and f only
- Standing eye heights of women normally distributed with a mean of 1516 mm and a standard deviation of 63 mm. What % of women have a standing eye height between 1420 mm and 1560 mm, and what is the probability that a group of 20 women has an average standing eye height that is less than 1500 mm? Even though the sample size is less than 30, why can the Central Limit Theorem still be applied?a) Find such that P( X < ? ) = 0.9332, where X is a normal random variable with mean = 10 and standard deviation = 2.5. b) Find such that P( X > ?) = 0.1230, where X is a normal random variable with mean = 10 and standard deviation = 2.5.Suppose X is a non-standard normal variable, with a mean of 118, and std deviation of 15 (i.e. X~N(118,15)). You have an observation (z) on a standard normal scale (i.e. Z~N(0,1)). If z = -0.03, what is it on an X scale?
- Assume that the random variable X is normally distributed with mean = 15 and standard deviation = 2. Let n = 4. Find P( > 16) and P( < 16).i need help with cYou are hired by the governor to study whether a tax on liquor has decreased average liquor consump- tion in your state. You are able to obtain, for a sample of individuals selected at random, the difference in liquor consumption (in ounces) for the years before and after the tax. For person i who is sampled randomly from the population, Y, denotes the change in liquor consumption. Treat these as a random sample from a Normal(µ, o²) distribution. (i) The null hypothesis is that there was no change in average liquor consumption. State this formally in terms of µ. (ii) The alternative is that there was a decline in liquor consumption; state the alternative in terms of µ. (iii) Now, suppose your sample size is n = 900 and you obtain the estimates ỹ = -32.8 and s = 466.4. Calculate the t statistic for testing Ho against H¡; obtain the p-value for the test. (Because of the large sample size, just use the standard normal distribution tabulated in Table G.1.) Do you reject H, at the 5% level?…