T-Distribution 5. Statistical Abstracts (117th edition) reports that the average amount spent annually for food by householders under 25 years of age is $2,690. A random sample of 16 people under 25 years of age who live in a university neighborhood were surveyed. The survey showed that they spent a sample mean amount of $3,220 with sample standard deviation $750. Test the claim that the mean for this neighborhood is different from the national average. Use a 5% significance level. Assume our data is normally distributed. State the null and alternate hypotheses. a. b. C. Would you perform a one-tailed or two-tailed test? Why? State the test statistic and p-value for your test. Round your answers to four decimal places.

Nutrition Iron-deficiency anemia is an important nutritional health problem in the United States. A dietary assessment was performed on 56 boys 9 to 11 years of age whose families were below the poverty level. The mean daily iron intake among these boys was found to be 12.40 mg with standard deviation 4.71 mg. Suppose the mean daily iron intake among a large population of 9- to 11-year-old boys from all income strata is 14.41 mg. We want to test whether the mean iron intake among the low-income group is different from that of the general population. You can use the Inferential Statistics page and the Distribution Calculators page in SALT to answer parts of this question. (a) State the hypotheses (in mg) that we can use to consider this question. (Enter != for as needed.) Ho: H₁: (b) Carry out the hypothesis test in part (a) using the critical-value method with an a level of 0.05, and summarize your findings. Find the test statistic. (Round your answer to two decimal places.) Find the…

You have have sampled five production lines operating and you wish to estimate the sample standard deviation of the five lines. Given the data below, provide your estimate of the sample standard deviation to the first decimal point, that is, 0.9 instead of 1. Month Production Line 1 51 Line 2 145 Line 3 70 Line 4 119 Line 5 92

Find out mean deviation from mean by applying direct method from the following distribution: Values: 48, 50, 46, 44, 40, 42, 35, 30, 20, 15.