Construct a 93% conidence interval for the population mean based on a sample of measuss 2.5, 7.4, 8.0, 7.4, 9.1 errors have Normal distribution, and the measurement device guarantees a standard deviatio
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A: given data normal distributionmean(μ) = 31.2standard deviation (σ) = 4.6for x = 39.5Z = ?
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A: Given n=32 x¯=53s=10α=0.05
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A: Null and alternative hypothesis: H0: μ = 5.6Ha: μ < 5.6
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A: Givensample size(n)=583significance level(α)=0.01
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A: From the given dataFor women's :Fro Men's :
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A: mean=μ=300, σ=sd=50X~N(300, 50)P(X<250or X>350)=?
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A: Answer: From the given data, mean(μ) = 160 standard deviation(σ) = 14
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A: Men mean =67.7 s = 3.5
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- A random sample of 100 pumpkins is obtained and the mean circumferenceis found to be 40.5 cm and standard deviation 1.6 cm. Use a 0.05 significance level to test the claim that the mean circumference of all pumpkins is equal to 39.9 cm.A survey found that women's heights are normally distributed with mean 63.3 in. and standard deviation 2.9 in. The survey also found that men's heights are normally distributed with mean 69.7 in. and standard deviation 3.6 in. Most of the live characters employed at an amusement park have height requirements of a minimum of 55 in. and a maximum of 64 in. What's the percentage of men meeting the height requirement. What does the result suggest about the genders of the people who are employed as characters at the amusement park?Total blood volume (in ml) per body weight is important in medical research. For healthy adults, the red blood cell volume mean is about 34 ml/kg. Red blood cell that is too low or too high can indicate a medical problem. Suppose that Roger has 20 blood tests and the mean is 35.4 ml/kg. Suppose that red blood cell volume follows normal distribution with standard deviation 2.34 ml/kg. With 5% significance level, test that the mean blood cell volume is different from 28ml/kg. Which test to use? a.) None of the answer choices b.) T-test for the Mean c.) T-test for Proportion d.) Z-test for the Mean e.) Z-test for Proportion
- Independent random samples of patients who had received knee and hip replacement were asked to assess the quality of service on a scale from 1 (low) to 7 (high). For a sample of 83 knee patients, the mean rating was 6.543 and the sample standard deviation was 0.649. For a sample of 54 hip patients, the mean rating was 6.733 and the sample standard deviation was 0.425. Test, against a two-sided alternative, the null hypothesis that the population mean ratings for these two types of patients are the same.A study was performed concerning risk factors for carotid-artery stenosis (arterial narrowing) among 264 men. The data reported is: Sample size Mean Standard deviationGroup 1: No stenosis 156 5.3 1.4Group 2: Stenosis 108 5.1 0.8 The test stastistic F is assumed to be 3.06 to check the equal variance. Question: Suppose researchers want to test if there is a significant difference in mean blood-glucose level between men with and without stenosis considering the critical F test is 1.35 with a equal variance (pooled), what is the test statistic for this test ?What are the limitations of a population range of 25, a population variance of 72.9167, and a population standard deviation of 8.5391?
- Suppose the total weight of passengers and luggage on an airline flight is normally distributed with mean 8914 and standard deviation 32 What is the 2.28th percentile of total passenger weightBig babies: The National Health Statistics Reports described a study in which a sample of 96 one-year-old baby boys were weighed. Their mean weight was 25.6 pounds with standard deviation 5.3 pounds. A pediatrician claims that the mean weight of one-year-old boys is greater than 25 pounds. Do the data provide convincing evidence that the pediatrician's claim is true? Use the =α0.05 level of significance and the P -value method and Excel. Part: 0 / 5 0 of 5 Parts Complete Part 1 of 5 (a)State the appropriate null and alternate hypotheses. H0 : =μ25 H1 : >μ25 This hypothesis test is a ▼two-tailed test. Part: 1 / 5 1 of 5 Parts Complete Part 2 of 5 (b)Compute the value of the test statistic. Round the answer to at least three decimal places.…A sports writer wants to see if a football filled with helium travels farther, on average, than a football filled with air. 12 footballs were filled with helium to the recommend pressure and 15 footballs were filled with air to the recommended pressure. The mean yardage for the helium filled footballs was 267 yards with a standard deviation of 3 yards. The mean yardage for the air filled footballs was 241 yards with a standard deviation of 5 yards. Assume the populations are normal with equal variances. (a). Construct a 99% confidence interval for the mean difference in in yardage for the two types of footballs Lower bound (use 3 decimal places) Upper bound (use 3 decimal places) (b). What can you conclude about the sports writer's idea that helium footballs travel farther, on average? The helium footballs are no different than the other footballs, on average The other footballs travel farther on average than the helium footballs The helium footballs travel farther on average than the…
- find and interpret the z-score for the darta value given/ the value 250 in a dataset with mean 208 amd standard deviation 21.Suppose the mean weight of King Penguins found in an Antarctic colony last year was 15.4 kg. In a sample of 35 penguins same time this year in the same colony, the mean penguin weight is 14.6 kg. Assume the population standard deviation is 2.5 kg. At .05 significance level, can we reject the null hypothesis that the mean penguin weight does not differ from last year? Use P-value approach.It appears that people who are mildly obese are less active than leaner people. One study looked at the average number of minutes per day that people spend standing or walking. Among mildly obese people, the mean number of minutes of daily activity (standing or walking) is approximately Normally distributed with 371 minutes and standard deviation 69 minutes. The mean number of minutes of daily activity for lean people is approximately Normally distributed with 526 minutes and standard deviation 106 minutes. A researcher records the minutes of activity for an SRS of 6 mildly obese people and an SRS of 6 lean people. (a) What is the probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes? (Enter your answer rounded to four decimal places.) probability: 0.0256 Incorrect (b) What is the probability that the mean numberpof minutes of daily activity of the 6 lean people exceeds 410 minutes? (Enter your answer rounded to four decimal…
