In a area there is a stationa for messuring air polution. It gived 5 readings every day. X1 , X2, ... Xn which are expected to be independent and identical normal distribuded with expectation µ and variance σ2,which are unknown. µ is the level of pollution. n= 5. and the following messurment are made: 252, 311, 268, 287, 302. Find a 95% confidence interval for µ based on the gived messurments. How do you interpurate a coverage of 95%
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In a area there is a stationa for messuring air polution. It gived 5 readings every day. X1 , X2, ... Xn which are expected to be independent and identical normal distribuded with expectation µ and variance σ2,which are unknown.
µ is the level of pollution. n= 5. and the following messurment are made: 252, 311, 268, 287, 302.
Find a 95% confidence interval for µ based on the gived messurments. How do you interpurate a coverage of 95%
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- An ecologist is studying the impact of local polluted waters on the growth of alligators. The length of adult male alligators typically follows a normal distribution with a standard deviation of 2 feet. The ecologist wants to estimate the mean length of this population of alligators. Suppose she samples n alligators at random and uses the sample mean, X to as an estimator for u. a. What is the bias and variance of the estimator? (Note, these may be a function of n.) b. If n = 4, what is the probability that the estimator is within one foot of the true mean? (I.e. find P(|X – µ| < 1). c. What sample size, n, is required for the estimator to be within one foot of the true mean with 95% probability? (I.e. find the value of n that satisfies P(|X – µ| < 1) = 0.95.) d. Suppose the ecologist ends up sampling n = 9 alligators and calculates a sample mean of ī = 10.4 feet. Construct a 95% confidence interval for the population mean. e. Give an interpretation for the interval obtained in (d).The mean IQ for the entire population in any age group is known to be 100 with a standard deviation of 15. Suppose the IQ scores of 72 12th graders in Columbus are measured and the sample mean is found to be 103.4 and the sample standard deviation is 20.2. Do the scores provide good evidence that the mean IQ of this population is not 100? State the null (H0) and alternative (Ha) hypotheses. 2. What are the confidence intervals for this data? 3. What is the value of the test statistic? 4. What is the p-value for this data? 5. What is your decision about the null hypothesis?: Reject Fail to Reject 6. Summarize the conclusion in the context of the problem.The following data represent the length of time, in days, to recovery for patients randomly treated with one of two medications to clear up severe bladder infections. Find a 99% confidence interval for the difference µ, - H, between in the mean recovery times for the two medications, assuming normal populations with equal variances. Medication 1 n = = 10 x, = 17 = 1.9 Medication 2 n2 = 14 X2 = Click here to view page 1 of the table of critical values of the t-distribution. Click here to view page 2 of the table of critical values of the t-distribution. = 19 S2 = 1.5 .... . The confidence interval isSuppose that the height of Asian males aged 35-44 in the United States is normally distributed with a standard deviation of 3 in. Jeremy takes a simple random sample of 16 such men and notices that none of them are unusually tall or unusually short. He then calculates the average of the sample to be 67 in. and is planning on using this to find a 90% z-confidence interval for the true mean height. Can Jeremy use this information to find the z-confidence interval? Complete the following sentences. The population distribution is and the population standard deviation is Therefore, the sample to find a 90% z-confidence interval.The ages of registered voters in Smith County are normally distributed with a population standard deviation of 3 years and an unknown population mean. A random sample of 18 voters is taken and results in a sample mean of 55 years. Find the margin of error for a 95% confidence interval for the population mean. z0.10z0.10 z0.05z0.05 z0.025z0.025 z0.01z0.01 z0.005z0.005 1.282 1.645 1.960 2.326 2.576 You may use a calculator or the common z values above. Round the final answer to two decimal places.Suppose X1,..., Xn is a random sample from an exponential distribution with mean e. If X = 17.9 with n = 50, find (a) a one-sided 95% confidence interval for 0, and (b) a two-sided 95% confidence interval for 0.A sample of n=41n=41 data values randomly selected from a normally distributed population has variance s2=30s2=30. Construct a 95% confidence interval for the population variance. Round your endpoints to one decimal place.Can you solve this problem about the confidence interval?A sample of 14 male college students consumes an average of 14.3 gallons of beer per month, with a sample standard deviation of 5.2 gallons. You can safely assume that the population of monthly beer consumption is Normally distributed. What is the margin of error for a 90% Confidence Interval for the mean? (Leave your answer to two decimal places, ex: 18.12)The following data represent the length of time, in days, to recovery for patients randomly treated with one of two medications to ciear up severe bladder infections. Find a 95% confidence interval for the difference u,-H, between in the mean recovery times for the two medications, assuming normal populations with equal variances. Medication 1 x, 13 s = 14 s =1.6 n, = 14 Medication 2 n211 X2= 16 Click here to view page 1 of the table of critical values of the t-distribution. Click here to view page 2 of the table ofcritical values of the t-distribution The confidence interval isThe following data represent the length of time, in days, to recovery for patients randomly treated with one of two medications to clear up severe bladder infections. Find a 90% confidence interval for the difference μ₂-μ₁ between in the mean recovery times for the two medications, assuming normal populations with equal variances. n₁ = 13 X₁ = 11 Medication 1 + X₂ = 16 Click here to view page 1 of the table of critical values of the t-distribution. Click here to view page 2 of the table of critical values of the t-distribution. Medication 2 = 19 n₂ = The confidence interval isThe commute times for the workers in a city are normally distributed with an unknown population mean and standard deviation. 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