Glucose Level: Horses –Gentle Ben is a Morgan horse at a Colorado dude ranch. Over the past 8 weeks, a veterinarian took the following glucose readings from this horse (in mg/100 ml).938882105991108489The sample mean is x̄= 93.8. Let x be a random variable representing glucose readings taken from Gentle Ben. We may assume that x has a normal distribution, and we know from experience that σ = 12.5. The mean glucose level for the horse should be μ = 85. Does the data indicate that Gentle Ben has an overall glucose level higher than 85? Use α = 0.05. l)Compute the test statistic since σ is known: m)Find the P-Value associated with this z score and the indicated test. n)Determine if the P-Value is greater or less than the level of significance (α).o)Do you reject or fail to reject the null hypothesis? p)Is there sufficient or insufficient evidence to reject or fail to reject the claim that Gentle Ben’s average glucose level is higher than average?

Women are recommended to consume 1700 calories per day. You suspect that the average calorie intake is different for women at your college. The data for the 12 women who participated in the study is shown below: 1784, 1677, 1583, 1984, 1910, 1737, 1522, 1574, 1853, 1546, 1707, 1866 Assuming that the distribution is normal, what can be concluded at the αα = 0.10 level of significance? For this study, we should use     The null and alternative hypotheses would be:       H0:H0:                  H1:H1:              The test statistic     = (please show your answer to 3 decimal places.)

Listed below are systolic blood pressure measurements​ (mm Hg) taken from the right and left arms of the same woman. Assume that the paired sample data is a simple random sample and that the differences have a distribution that is approximately normal. Use a 0.10significance level to test for a difference between the measurements from the two arms. What can be​ concluded?   Right arm 146 137 140 132 132   Left-arm 183 172 179 156 149   In this​ example, μd is the mean value of the differences d for the population of all pairs of​ data, where each individual difference d is defined as the measurement from the right arm minus the measurement from the left arm. What are the null and alternative hypotheses for the hypothesis test?   Identify the test statistic.   Identify the​ P-value.

Our environment is very sensitive to the amount of ozone in the upper atmosphere. The level of ozone normally found is 7.97.9 parts/million (ppm). A researcher believes that the current ozone level is at an excess level. The mean of 2424 samples is 8.18.1 ppm with a variance of 0.250.25. Does the data support the claim at the 0.10.1 level? Assume the population distribution is approximately normal. Step 3 of 5:  Specify if the test is one-tailed or two-tailed.

Assume that you have been provided the following information for a sample of cars The covariance between weight and miles per gallon is equal to -4.9 pound-miles. The standard deviation of the cars' mileages is 7.75. The standard deviation of the cars' weights is 0.94. Estimate the correlation coefficient to the first decimal place.

The Department of Agriculture wants to determine whether the mean yield per acre for a particular variety of soybeans is different this year compared to the historical avg yield of 520 bushels per acre. Thus, the null and alternative hypotheses are  Ho: mu = 520 Ha: mu is not = to 520 You set out to conduct a hypothesis test at the 95% level. Mean yields are distributed Normally. You take a sample of 36 different acres. From this sample, you calculate a mean yield of 573 and a std deviation of 24.  Do you reject or not-reject the null hypothesis? Group of answer choices Do not reject the null hypothesis. Reject the null hypothesis.

Our environment is very sensitive to the amount of ozone in the upper atmosphere. The level of ozone normally found is 7.9 parts/million (ppm). A researcher believes that the current ozone level is at an excess level. The mean of 24 samples is 8.1 ppm with a variance of 0.25. Does the data support the claim at the 0.1 level? Assume the population distribution is approximately normal. Step 5 of 5:  Make the decision to reject or fail to reject the null hypothesis.

Our environment is very sensitive to the amount of ozone in the upper atmosphere. The level of ozone normally found is 4.6 parts/million (ppm). A researcher believes that the current ozone level is at an excess level. The mean of 14 samples is 4.9 ppm with a variance of 1.2 Does the data support the claim at the 0.01 level? Assume the population distribution is approximately normal. Step 2 of 5 :   Find the value of the test statistic. Round your answer to three decimal places.

Women are recommended to consume 1700 calories per day. You suspect that the average calorie intake is different for women at your college. The data for the 12 women who participated in the study is shown below: 1808, 1681, 1723, 1847, 1808, 1858, 1461, 1449, 1631, 1635, 1815, 1874 Assuming that the distribution is normal, what can be concluded at the αα = 0.01 level of significance? For this study, we should use     The null and alternative hypotheses would be:       H0:H0:                   H1:H1:               The test statistic     =  (please show your answer to 3 decimal places.) The p-value =  (Please show your answer to 4 decimal places.) The p-value is     αα Based on this, we should      the null hypothesis. Thus, the final conclusion is that ... The data suggest the population mean is not significantly different from 1700 at αα = 0.01, so there is sufficient evidence to conclude that the population mean calorie intake for women at your college is equal to 1700. The data…

Refer to the below table. Using an alpha = 0.05, test the claim that IQ scores are the same for children in three different blood lead level groups: low lead level, medium lead level, and high lead level).   One-Way Analysis of Variance Summary Table for IQ Measurements for Children among Three Blood Lead Level Groups: Low Lead Level, Medium Lead Level, and High Lead Level. Source df SS MS F p             Between-group (treatment) 2   469.1827 2677.864 2.30 0.104 Within-group (error)       118    203.6918 11745.05     Total       120