A prediction model was developed to predict the calories based on the total amount of fat in a food item. The prediction model is as follows: predicted calories = 134 + 22.7 * total_fat ± 44 Using the prediction model given above, a fast food item with total_fat = 9.5 is predicted to have a calories of between 393.7 1532.7 [Select] 305.7 -37.7 [Select] 305.7 393.7 -125.7 1444.7 and If the actual calories of this fast food item was 346.6, that would mean that the prediction interval generated from this model was [Select] correct incorrectl
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Pick the correct answers and tell me why and how
![A prediction model was developed to predict the calories based on the total
amount of fat in a food item. The prediction model is as follows:
predicted calories = 134 + 22.7 * total_fat ± 44
Using the prediction model given above, a fast food item with total_fat = 9.5 is
predicted to have a calories of between [Select]
393.7
1532.7
[Select]
305.7
-37.7
305.7
393.7
-125.7
1444.7
and
If the actual calories of this fast food item was 346.6, that would mean that the
prediction interval generated from this model was [Select]
correct
incorrectl](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F647d9ea8-515f-42e1-b0fd-52a2f87df7ff%2F8486b4da-f393-469c-b9d4-20c751e8be6e%2F6tms14q_processed.png&w=3840&q=75)
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- According to the National Health and Nutrition Examination Survey (NHANES) sponsored by the U.S. government, a random sample of 712 males between 20 and 29 years of age and a random sample of 1,001 males over the age of 75 were chosen and the weight of each of the males were recorded (in kg). Do the data provide evidence that the younger male population weighs more (on average) than the older male population? Use “Y” for ages 20-29 and “S” for ages 75+. It was found that x̅Y=83.4, sY=18.7, x̅S=78.5, and sS=19.0. A) What are the correct null and alternative hypotheses? B)What is the parameter of interest? The difference between the mean weight of all males ages 20-29 and the mean weight of all males ages 75+ The mean weight (in kg) The difference between the mean weight of the sample of males ages 20-29 and the mean weight of the sample of males ages 75+ The mean difference between the weights of males ages 20-29 and males ages 75+ c)Is the randomness condition met for…According to the National Health and Nutrition Examination Survey (NHANES) sponsored by the U.S. government, a random sample of 712 males between 20 and 29 years of age and a random sample of 1,001 males over the age of 75 were chosen and the weight of each of the males were recorded (in kg). Do the data provide evidence that the younger male population weighs more (on average) than the older male population? Use “Y” for ages 20-29 and “S” for ages 75+. It was found that x̅Y=83.4, sY=18.7, x̅S=78.5, and sS=19.0. a) Is the normality condition met for this test? (see picture) No, since the histograms show skewness and outliers . Yes, since both sample sizes are larger than 30. No, since both sample sizes, nY and nS, are less than 30. Yes, since it was stated that the two populations are normally distributed. b)What are the correct degrees of freedom for this test? Group of answer choices 999 1000 1712 1711 711According to the National Health and Nutrition Examination Survey (NHANES) sponsored by the U.S. government, a random sample of 712 males between 20 and 29 years of age and a random sample of 1,001 males over the age of 75 were chosen and the weight of each of the males were recorded (in kg). Do the data provide evidence that the younger male population weighs more (on average) than the older male population? Use “Y” for ages 20-29 and “S” for ages 75+. It was found that x̅Y=83.4, sY=18.7, x̅S=78.5, and sS=19.0. a)Suppose the test statistic is t = 2.398. What is the associated p-value? Group of answer choices 0.001 < p-value < 0.002 0.005 < p-value < 0.01 0.01 < p-value < 0.02 0.0005 < p-value < 0.001 b) Suppose the p-value is 0.02 < p-value < 0.04. At α = 0.10 what is the appropriate conclusion to make? Group of answer choices Fail to reject H0 and conclude that the mean weight of all males ages 20-29 is greater than the mean weight of all…
- what is the value of 1.1538 from the z_table?The Us food and Drug administration has a limit for mercury content in fish of 1.0ppm (parts per million) while in Canada the limit is 0.5 ppm. Using the variable average mercury in the Florida lakes dataset to test whether there is evidence that average mercury level of fish ( large mouth Bass) in Florida lakes is less than 1.0ppm and whether there is evidence that average mercury level of fish in Florida lakes is less than 0.5ppm. STATE THE NULL AND ALTERNATIVE HYPOTHESES? GIVE THE TEST STATISTICS AND P-VALUE? round the test statistic to two decimal places and the p-value to three decimal places. DOES THIS SAMPLE PROVIDE EVIDENCE THAT THE AVERAGE MERCURY LEVEL OF FISH (LARGE MOUTH BASS) IN FLORIDA LAKES IS LESS THAN 1.0PPM? LESS THAN 0.5 PPM : STATE THE NULL AND ALTERNATIVE HYPOTHESES. GIVE THE TEST STATISTICS AND P-VALUE ? round the test statistic to two decimal places and the p-value to three decimal places. DOES THIS SAMPLE PROVIDE EVIDENCE THAT THE AVERAGE MERCURY…The Southern Luzon Expressway was designed to carry a free-flow speed of 50kph and a capacity of 4000 vehicles per hour. At a certain point near theintersection of a road leading to Laguna there were 3600 vehicles counted,what is the space mean speed of these vehicles in kph?
- Industrial light bulbs should have a mean life length acceptable to potential users and a relatively small variation in life length. If some bulbs fail too early in their life, users become annoyed and are likely to switch to bulbs produced by a different manufacturer. Large variations above the mean reduce replacement sales; in general, variation in life lengths disrupts the user’s replacement schedules. A random sample of 20 bulbs produced by a particular manufacturer produced the following lengths of life (in hours):2100, 2302, 1951, 2067, 2415, 1883, 2101, 2146, 2278, 2019, 1924, 2183, 2077, 2392, 2286, 2501, 1946, 2161, 2253, 1827.(a) Find a 99% confidence interval for the standard deviation of the lengths of life for the bulbs produced by this manufacturer. (b) Is the true population standard deviation possibly less than 150 hours? Why or why not?One tree in the study had a weight that exceeded its expected weight (according to the best fit curve) by more than that of any other tree. By how many tons did the trees actual weight exceed the expected weight for its height?A magizine claimed that more than 55% of adults skip breakfast at least 3 days a week. To test this, a dietician selected a random sample of 80 adults and asked them how many days a week they skip breakfast. 45 of them responded that they skip breakfast at least 3 days a week. At a= .10, test the magazine's claim
- Obesity is an increasing problem in America among all age groups. A recent study reported 1276 individuals in a sample af 4115 adults were found to be obese (a bady mass index exceeding 30; this index is a measure of weight relative ta height). Five years ago, survey based on peaple's own assessmentrevealed that 20% of aduit Americans considered themselves obese. Does the mastrecent data suggest that the true proportion of adults who are cbese is more than 1.5 times the percentage from the self-assessment survey?Carry out a test using a significance level of 5%. Question 1 State your hypotheses. Họ:p = 1.5 vs HA:p 1.5 O Ho:p= po Vs HA:p> 1.5po O Hoip-0.3 vs Ha : pr 0.3 O Ho:p= 0.2 vs Ha p> 0.2 O Ho:p =0.2 vs HA : p 0.2 O Hoip=0.3 vs HA : p> 0.3A sample of 100 bulbs of brand A gave a mean lifetime of 1200 h, with an SD of 70 h, while another sample of 120 bulbs of brand B gave a mean lifetime of 1150 h with a SD of 85 h. Can we conchude that brand A bulbs are superior to brand B bulbs at 1% LOS?The average time period that spouses survive after the death of their partner is said to be greater for widows than for widowers. A sociologist who wishes to evaluate such claim obtained survival data (in years) following the death of a spouse on randomly selected 10 widows and 8 widowers. Software Output: Shapiro-Wilk TestFor Widows: data: survival p-value = 0.57For Widowers: data: survival p-value = 0.98Levene’s Test for Homogeneity of Variance p-value = 0.98 What is the most appropriate test procedure to test the claim? a. Z-test b. Pooled t-test c. Non-pooled t-test d. Mann-Whitney Test